From Shang Gao Magic 15 If a 2 + b 2 = c 2 , then ( a , b , c ) is - - PowerPoint PPT Presentation

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Magic 15

Magic 15: A STORY OVER 380 YEARS

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

From Shang Gao

If a2 + b2 = c2 , then (a, b, c) is called to be a Pythagorean triple. (3, 4, 5), (5, 12, 13) are Pythagorean

• triples. If Pythagorean triple (a, b, c) satisfies

gcd(a, b, c) = 1, then it is primitive.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

From Shang Gao

If a2 + b2 = c2 , then (a, b, c) is called to be a Pythagorean triple. (3, 4, 5), (5, 12, 13) are Pythagorean

• triples. If Pythagorean triple (a, b, c) satisfies

gcd(a, b, c) = 1, then it is primitive. (3, 4, 5), (5, 12, 13) are primitive, but (6, 8, 10) is not.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

From Shang Gao

If a2 + b2 = c2 , then (a, b, c) is called to be a Pythagorean triple. (3, 4, 5), (5, 12, 13) are Pythagorean

• triples. If Pythagorean triple (a, b, c) satisfies

gcd(a, b, c) = 1, then it is primitive. (3, 4, 5), (5, 12, 13) are primitive, but (6, 8, 10) is not. Question 1. How many primitive Pythagorean triples can you find? What is the largest one?

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

From Shang Gao

If a2 + b2 = c2 , then (a, b, c) is called to be a Pythagorean triple. (3, 4, 5), (5, 12, 13) are Pythagorean

• triples. If Pythagorean triple (a, b, c) satisfies

gcd(a, b, c) = 1, then it is primitive. (3, 4, 5), (5, 12, 13) are primitive, but (6, 8, 10) is not. Question 1. How many primitive Pythagorean triples can you find? What is the largest one? Answer: Infinitely many (so no ”Largest” can exist), since there are infinitely many rational points on x2 + y2 = 1 (via parametrization using the line x

t + y = 1).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2, (x, y) ∈ Z2?

Two Square Theorem. (1640, Fermat-1747, Euler) If p is a prime, then p = x2 + y2 ⇐ ⇒ p ≡ 1(mod4).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2, (x, y) ∈ Z2?

Two Square Theorem. (1640, Fermat-1747, Euler) If p is a prime, then p = x2 + y2 ⇐ ⇒ p ≡ 1(mod4).

• Corollary. n = x2 + y2 ⇐

⇒ n has no prime divisor of the form 4k + 3.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2, (x, y) ∈ Z2?

Two Square Theorem. (1640, Fermat-1747, Euler) If p is a prime, then p = x2 + y2 ⇐ ⇒ p ≡ 1(mod4).

• Corollary. n = x2 + y2 ⇐

⇒ n has no prime divisor of the form 4k + 3. Euler’s proof needs 5 lemmas, totally about 5 pages.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Dedekind’s Proof of Two Square Theorem

In 1894, Dedekind gave a proof of Two Square Theorem using Gauss integer ring Z[i] = {a + bi | a, b ∈ Z} where i = √−1 and its quotient Z[i]/(p) with p a prime number.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Dedekind’s Proof of Two Square Theorem

In 1894, Dedekind gave a proof of Two Square Theorem using Gauss integer ring Z[i] = {a + bi | a, b ∈ Z} where i = √−1 and its quotient Z[i]/(p) with p a prime number. What’s Z[i]/(p)? Or easier, what’s Z/(p)?

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Z/(p) = {0, 1, 2, · · · , p − 1} with natural ”+” and ”×” is a finite field, called Galois finite field and denoted by GF(p) (or Fp), where the capital letter ”G” is E. Galois(1811-1831).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Z/(p) = {0, 1, 2, · · · , p − 1} with natural ”+” and ”×” is a finite field, called Galois finite field and denoted by GF(p) (or Fp), where the capital letter ”G” is E. Galois(1811-1831). When p = 2, Z/(2) = {0, 1} has the following ”+” and ”×” table:

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Z/(p) = {0, 1, 2, · · · , p − 1} with natural ”+” and ”×” is a finite field, called Galois finite field and denoted by GF(p) (or Fp), where the capital letter ”G” is E. Galois(1811-1831). When p = 2, Z/(2) = {0, 1} has the following ”+” and ”×” table: + 1 1 1 1 × 1 1 1

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Z/(p) = {0, 1, 2, · · · , p − 1} with natural ”+” and ”×” is a finite field, called Galois finite field and denoted by GF(p) (or Fp), where the capital letter ”G” is E. Galois(1811-1831). When p = 2, Z/(2) = {0, 1} has the following ”+” and ”×” table: + 1 1 1 1 × 1 1 1 The above tables tell us exactly what parity is!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Z/(p) = {0, 1, 2, · · · , p − 1} with natural ”+” and ”×” is a finite field, called Galois finite field and denoted by GF(p) (or Fp), where the capital letter ”G” is E. Galois(1811-1831). When p = 2, Z/(2) = {0, 1} has the following ”+” and ”×” table: + 1 1 1 1 × 1 1 1 The above tables tell us exactly what parity is! Can you partition all rational numbers into odd ones and even ones? Real or complex numbers?Can you partition all integral polynomials into odd ones and even ones?

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Elementary Geometry over F2

What are the straight lines on the plane F 2

2 ?

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Elementary Geometry over F2

What are the straight lines on the plane F 2

2 ?

Answer: Easy. Totally 6 (2 points = 1 line): x = 0, x = 1, y = 0, y = 1, x + y = 0, x + y = 1.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Elementary Geometry over F2

What are the straight lines on the plane F 2

2 ?

Answer: Easy. Totally 6 (2 points = 1 line): x = 0, x = 1, y = 0, y = 1, x + y = 0, x + y = 1. What are the circles on the plane F2? How many are they?

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Elementary Geometry over F2

What are the straight lines on the plane F 2

2 ?

Answer: Easy. Totally 6 (2 points = 1 line): x = 0, x = 1, y = 0, y = 1, x + y = 0, x + y = 1. What are the circles on the plane F2? How many are they? Answer: A little surprising. Only 2. ( Comparing: ”More” circles than lines in R2.)

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Elementary Geometry over F2

What are the straight lines on the plane F 2

2 ?

Answer: Easy. Totally 6 (2 points = 1 line): x = 0, x = 1, y = 0, y = 1, x + y = 0, x + y = 1. What are the circles on the plane F2? How many are they? Answer: A little surprising. Only 2. ( Comparing: ”More” circles than lines in R2.) More surprising? All circles are straight lines! (Comparing: Lines on the sphere S2 are circles)! They are: x + y = 0 and x + y = 1

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Elementary Geometry over F2

What are the straight lines on the plane F 2

2 ?

Answer: Easy. Totally 6 (2 points = 1 line): x = 0, x = 1, y = 0, y = 1, x + y = 0, x + y = 1. What are the circles on the plane F2? How many are they? Answer: A little surprising. Only 2. ( Comparing: ”More” circles than lines in R2.) More surprising? All circles are straight lines! (Comparing: Lines on the sphere S2 are circles)! They are: x + y = 0 and x + y = 1 Much more surprising? Every point in F 2

2 is the center of

both circles of x + y = 0 and x + y = 1!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Elementary Geometry over F2

What are the straight lines on the plane F 2

2 ?

Answer: Easy. Totally 6 (2 points = 1 line): x = 0, x = 1, y = 0, y = 1, x + y = 0, x + y = 1. What are the circles on the plane F2? How many are they? Answer: A little surprising. Only 2. ( Comparing: ”More” circles than lines in R2.) More surprising? All circles are straight lines! (Comparing: Lines on the sphere S2 are circles)! They are: x + y = 0 and x + y = 1 Much more surprising? Every point in F 2

2 is the center of

both circles of x + y = 0 and x + y = 1! Much more surprising again? Every number in F2 can be the radius of both circles of x + y = 0 and x + y = 1!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Reason: (a + b)2 = a2 + b2 in F2–Pupils’ Dream!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Reason: (a + b)2 = a2 + b2 in F2–Pupils’ Dream! In Fp: (a + b)p = ap + bp, ap = a (Fermat’s Little Theorem & Gauss’ Primitive Root Theorem)!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Reason: (a + b)2 = a2 + b2 in F2–Pupils’ Dream! In Fp: (a + b)p = ap + bp, ap = a (Fermat’s Little Theorem & Gauss’ Primitive Root Theorem)! Z[i]/(p) ≃ Z/(p) ⊕ Z/(p)[i], so |Z[i]/(p)| = p2.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Reason: (a + b)2 = a2 + b2 in F2–Pupils’ Dream! In Fp: (a + b)p = ap + bp, ap = a (Fermat’s Little Theorem & Gauss’ Primitive Root Theorem)! Z[i]/(p) ≃ Z/(p) ⊕ Z/(p)[i], so |Z[i]/(p)| = p2. Recall 1. Z[i] is a Euclidean domain with Euclidean function N(x + iy) = x2 + y2, N(ab) = N(a)N(b).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Reason: (a + b)2 = a2 + b2 in F2–Pupils’ Dream! In Fp: (a + b)p = ap + bp, ap = a (Fermat’s Little Theorem & Gauss’ Primitive Root Theorem)! Z[i]/(p) ≃ Z/(p) ⊕ Z/(p)[i], so |Z[i]/(p)| = p2. Recall 1. Z[i] is a Euclidean domain with Euclidean function N(x + iy) = x2 + y2, N(ab) = N(a)N(b). Recall 2. Every Euclidean domain is a principal ideal ring.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Galois Finite Fields

Reason: (a + b)2 = a2 + b2 in F2–Pupils’ Dream! In Fp: (a + b)p = ap + bp, ap = a (Fermat’s Little Theorem & Gauss’ Primitive Root Theorem)! Z[i]/(p) ≃ Z/(p) ⊕ Z/(p)[i], so |Z[i]/(p)| = p2. Recall 1. Z[i] is a Euclidean domain with Euclidean function N(x + iy) = x2 + y2, N(ab) = N(a)N(b). Recall 2. Every Euclidean domain is a principal ideal ring.

• Conclusion. Z[i] is a principal ideal ring.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Dedekind’s Proof of Two Square Theorem(continued)

If p is a prime, then ip−1 = (−1)

p−1 2 . Let

w = x + iy ∈ Z[i]/(p)

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Dedekind’s Proof of Two Square Theorem(continued)

If p is a prime, then ip−1 = (−1)

p−1 2 . Let

w = x + iy ∈ Z[i]/(p) wp = xp + ipyp = xp + iyp(−1)

p−1 2

(modp).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Dedekind’s Proof of Two Square Theorem(continued)

If p is a prime, then ip−1 = (−1)

p−1 2 . Let

w = x + iy ∈ Z[i]/(p) wp = xp + ipyp = xp + iyp(−1)

p−1 2

(modp). If p ≡ 1(mod4), then (since xp = x for all x ∈ Z/(p)) wp = xp + iyp (modp) = x + iy (modp)

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Dedekind’s Proof of Two Square Theorem(continued)

If p is a prime, then ip−1 = (−1)

p−1 2 . Let

w = x + iy ∈ Z[i]/(p) wp = xp + ipyp = xp + iyp(−1)

p−1 2

(modp). If p ≡ 1(mod4), then (since xp = x for all x ∈ Z/(p)) wp = xp + iyp (modp) = x + iy (modp) ⇒Frobenius automorphism F : w → wp is identity.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Dedekind’s Proof of Two Square Theorem(continued)

If p is a prime, then ip−1 = (−1)

p−1 2 . Let

w = x + iy ∈ Z[i]/(p) wp = xp + ipyp = xp + iyp(−1)

p−1 2

(modp). If p ≡ 1(mod4), then (since xp = x for all x ∈ Z/(p)) wp = xp + iyp (modp) = x + iy (modp) ⇒Frobenius automorphism F : w → wp is identity. Kummer’s Theorem. Let O(F) be the order of F. If O(F) ∈ {1, 2}. Then the ideal (p) in Z[i] is a product of

2 O(F) distinct prime ideals.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Dedekind’s Proof of Two Square Theorem(continued)

If p is a prime, then ip−1 = (−1)

p−1 2 . Let

w = x + iy ∈ Z[i]/(p) wp = xp + ipyp = xp + iyp(−1)

p−1 2

(modp). If p ≡ 1(mod4), then (since xp = x for all x ∈ Z/(p)) wp = xp + iyp (modp) = x + iy (modp) ⇒Frobenius automorphism F : w → wp is identity. Kummer’s Theorem. Let O(F) be the order of F. If O(F) ∈ {1, 2}. Then the ideal (p) in Z[i] is a product of

2 O(F) distinct prime ideals.

So (p) = IJ = (a + bi)(c + di).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Dedekind’s Proof of Two Square Theorem(continued)

If p is a prime, then ip−1 = (−1)

p−1 2 . Let

w = x + iy ∈ Z[i]/(p) wp = xp + ipyp = xp + iyp(−1)

p−1 2

(modp). If p ≡ 1(mod4), then (since xp = x for all x ∈ Z/(p)) wp = xp + iyp (modp) = x + iy (modp) ⇒Frobenius automorphism F : w → wp is identity. Kummer’s Theorem. Let O(F) be the order of F. If O(F) ∈ {1, 2}. Then the ideal (p) in Z[i] is a product of

2 O(F) distinct prime ideals.

So (p) = IJ = (a + bi)(c + di). p = (a + bi)(c + di) ⇒ p2 = (a2 + b2)(c2 + d2) ⇒ p = a2 + b2 or p = c2 + d2. Done!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

One Sentence Proof of Two Square Theorem

One Sentence Proof. (1990, Don Zagier) Let p = 4k + 1 be a prime, then the set S = {(x, y, z) ∈ N3 | x2 + 4yz = p} is finite.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

One Sentence Proof of Two Square Theorem

One Sentence Proof. (1990, Don Zagier) Let p = 4k + 1 be a prime, then the set S = {(x, y, z) ∈ N3 | x2 + 4yz = p} is finite. S has an obvious involution f : (x, y, z) → (x, z, y).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

One Sentence Proof of Two Square Theorem

One Sentence Proof. (1990, Don Zagier) Let p = 4k + 1 be a prime, then the set S = {(x, y, z) ∈ N3 | x2 + 4yz = p} is finite. S has an obvious involution f : (x, y, z) → (x, z, y). If f has a fixed point (x, y, y), then p = x2 + (2y)2, which will prove the Two Square Theorem.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

One Sentence Proof of Two Square Theorem

One Sentence Proof. (1990, Don Zagier) Let p = 4k + 1 be a prime, then the set S = {(x, y, z) ∈ N3 | x2 + 4yz = p} is finite. S has an obvious involution f : (x, y, z) → (x, z, y). If f has a fixed point (x, y, y), then p = x2 + (2y)2, which will prove the Two Square Theorem.

• Observation. |S| has the same parity with the number of

fixed points of any involutions.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

One Sentence Proof of Two Square Theorem

One Sentence Proof. (1990, Don Zagier) Let p = 4k + 1 be a prime, then the set S = {(x, y, z) ∈ N3 | x2 + 4yz = p} is finite. S has an obvious involution f : (x, y, z) → (x, z, y). If f has a fixed point (x, y, y), then p = x2 + (2y)2, which will prove the Two Square Theorem.

• Observation. |S| has the same parity with the number of

fixed points of any involutions. Zagier’s idea. Find another involution having at least one fixed point.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

One Sentence Proof of Two Square Theorem(continued)

Zagier’s involution over S is a little difficult.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

One Sentence Proof of Two Square Theorem(continued)

Zagier’s involution over S is a little difficult. g : (x, y, z) →    (x + 2z, z, y − x − z) if x < y − z (2y − x, y, x − y + z) if y − z < x < 2y (x − 2y, x − y + z, y) if x > 2y

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

One Sentence Proof of Two Square Theorem(continued)

Zagier’s involution over S is a little difficult. g : (x, y, z) →    (x + 2z, z, y − x − z) if x < y − z (2y − x, y, x − y + z) if y − z < x < 2y (x − 2y, x − y + z, y) if x > 2y Easy to check g is an involution with unique fixed point (1, 1, k).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

One Sentence Proof of Two Square Theorem(continued)

Zagier’s involution over S is a little difficult. g : (x, y, z) →    (x + 2z, z, y − x − z) if x < y − z (2y − x, y, x − y + z) if y − z < x < 2y (x − 2y, x − y + z, y) if x > 2y Easy to check g is an involution with unique fixed point (1, 1, k). So the number of fixed points of f is odd and f has at least one fixed point (a, b, c) with b = c. Therefore p = a2 + 4bc = a2 + (2b)2. Done!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2 + z2, (x, y, z) ∈ Z3?

num = ∆ + ∆ + ∆! —Gauss’ Diary—This is Fermat’s Triangular Theorem.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2 + z2, (x, y, z) ∈ Z3?

num = ∆ + ∆ + ∆! —Gauss’ Diary—This is Fermat’s Triangular Theorem. n = ∆x + ∆y + ∆z Where ∆x = x(x+1)

2

is a ”triangular number” = 1 + 2 + · · · + x.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2 + z2, (x, y, z) ∈ Z3?

num = ∆ + ∆ + ∆! —Gauss’ Diary—This is Fermat’s Triangular Theorem. n = ∆x + ∆y + ∆z Where ∆x = x(x+1)

2

is a ”triangular number” = 1 + 2 + · · · + x. 8n + 3 = (2x + 1)2 + (2y + 1)2 + (2z + 1)2.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2 + z2, (x, y, z) ∈ Z3?

num = ∆ + ∆ + ∆! —Gauss’ Diary—This is Fermat’s Triangular Theorem. n = ∆x + ∆y + ∆z Where ∆x = x(x+1)

2

is a ”triangular number” = 1 + 2 + · · · + x. 8n + 3 = (2x + 1)2 + (2y + 1)2 + (2z + 1)2. Gauss is able to prove 3-Square Theorem.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Three Square Theorem

Three Square Theorem. (1798, Legendre) n = x2 + y2 + z2 ⇔ n = 4k(8m + 7).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Three Square Theorem

Three Square Theorem. (1798, Legendre) n = x2 + y2 + z2 ⇔ n = 4k(8m + 7). There are some stories between Gauss and Legendre.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Three Square Theorem

Three Square Theorem. (1798, Legendre) n = x2 + y2 + z2 ⇔ n = 4k(8m + 7). There are some stories between Gauss and Legendre. The first positive integer can’t be represented by a sum of three squares is 7.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2 + z2 + w 2, (x, y, z, w) ∈ Z4?

4-Square Theorem. (1640, Fermat-1770, Lagrange) For all n ∈ N, ∃(x, y, z, w) ∈ Z4, n = x2 + y2 + z2 + w2.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2 + z2 + w 2, (x, y, z, w) ∈ Z4?

4-Square Theorem. (1640, Fermat-1770, Lagrange) For all n ∈ N, ∃(x, y, z, w) ∈ Z4, n = x2 + y2 + z2 + w2. 1760, Euler’s Identity: (a2 +b2 +c2 +d2)(p2 +q2 +r2 +s2) = x2 +y2 +z2 +w2.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2 + z2 + w 2, (x, y, z, w) ∈ Z4?

4-Square Theorem. (1640, Fermat-1770, Lagrange) For all n ∈ N, ∃(x, y, z, w) ∈ Z4, n = x2 + y2 + z2 + w2. 1760, Euler’s Identity: (a2 +b2 +c2 +d2)(p2 +q2 +r2 +s2) = x2 +y2 +z2 +w2. So enough to prove 4-Square Theorem for all primes p = 4k + 3. One of the simplest way is again through Z[i]/(p). Other easy ways are through Hamilton quaternion or Minkowski geometry.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

What about x2 + y 2 + z2 + w 2, (x, y, z, w) ∈ Z4?

4-Square Theorem. (1640, Fermat-1770, Lagrange) For all n ∈ N, ∃(x, y, z, w) ∈ Z4, n = x2 + y2 + z2 + w2. 1760, Euler’s Identity: (a2 +b2 +c2 +d2)(p2 +q2 +r2 +s2) = x2 +y2 +z2 +w2. So enough to prove 4-Square Theorem for all primes p = 4k + 3. One of the simplest way is again through Z[i]/(p). Other easy ways are through Hamilton quaternion or Minkowski geometry. Write ax2 + by2 + cz2 + dw2 = [a; b; c; d].

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Universal Polynomials: Ramanujan’s List, 1916

Let X = (x1, · · · , xn). A quadratic form f (X) ∈ Z[X] is universal if it represents all positive integers.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Universal Polynomials: Ramanujan’s List, 1916

Let X = (x1, · · · , xn). A quadratic form f (X) ∈ Z[X] is universal if it represents all positive integers. [1;1;1;1], [1;1;1;2], [1;1;1;3], [1;1;1;4], [1;1;1;5], [1;1;1;6], [1;1;1;7], [1;1;2;2], [1;1;2;3], [1;1;2;4], [1;1;2;5], [1;1;2;6], [1;1;2;7], [1;1;2;8], [1;1;2;9], [1;1;2;10], [1;1;2;11], [1;1;2;12], [1;1;2;13], [1;1;2;14], [1;1;3;3], [1;1;3;4], [1;1;3;5], [1;1;3;6], [1;2;2;2], [1;2;2;3], [1;2;2;4], [1;2;2;5], [1;2;2;6], [1;2;2;7], [1;2;3;3], [1;2;3;4], [1;2;3;5], [1;2;3;6], [1;2;3;7], [1;2;3;8], [1;2;3;9], [1;2;3;10], [1;2;4;4], [1;2;4;5], [1;2;4;6], [1;2;4;7], [1;2;4;8], [1;2;4;9], [1;2;4;10], [1;2;4;11], [1;2;4;12], [1;2;4;13], [1;2;4;14], [1;2;5;5], [1;2;5;6], [1;2;5;7], [1;2;5;8], [1;2;5;9], [1;2;5;10]

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

A Story of Ramanujan and Universal Polynomials

Hardy went to see Ramanujan at a hospital, telling that his taxi number is 1729, a dull integer.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

A Story of Ramanujan and Universal Polynomials

Hardy went to see Ramanujan at a hospital, telling that his taxi number is 1729, a dull integer. No, Ramanujan replied quickly, 1729 is very interesting. It is the least positive integer that can be represented in two ways as sums of two cubes: 1729 = a3 + b3 = c3 + d3. Please give the 2 solutions.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

A Story of Ramanujan and Universal Polynomials

Hardy went to see Ramanujan at a hospital, telling that his taxi number is 1729, a dull integer. No, Ramanujan replied quickly, 1729 is very interesting. It is the least positive integer that can be represented in two ways as sums of two cubes: 1729 = a3 + b3 = c3 + d3. Please give the 2 solutions. So everyone believes Ramanujan’s list. In 1948, in his PhD Thesis, M.Willerding gave another list containing ”ALL” 178 general quadratic forms (with cross-terms).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

A Story of Ramanujan and Universal Polynomials

Hardy went to see Ramanujan at a hospital, telling that his taxi number is 1729, a dull integer. No, Ramanujan replied quickly, 1729 is very interesting. It is the least positive integer that can be represented in two ways as sums of two cubes: 1729 = a3 + b3 = c3 + d3. Please give the 2 solutions. So everyone believes Ramanujan’s list. In 1948, in his PhD Thesis, M.Willerding gave another list containing ”ALL” 178 general quadratic forms (with cross-terms). [1;2;5;5] is not in this list.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

A Story of Ramanujan and Universal Polynomials

Hardy went to see Ramanujan at a hospital, telling that his taxi number is 1729, a dull integer. No, Ramanujan replied quickly, 1729 is very interesting. It is the least positive integer that can be represented in two ways as sums of two cubes: 1729 = a3 + b3 = c3 + d3. Please give the 2 solutions. So everyone believes Ramanujan’s list. In 1948, in his PhD Thesis, M.Willerding gave another list containing ”ALL” 178 general quadratic forms (with cross-terms). [1;2;5;5] is not in this list. In 1993, J.Conway and his student W.Schneeberger gave a new list containing ”ALL” 204 such quadratic forms.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

A Story of Ramanujan and Universal Polynomials

Hardy went to see Ramanujan at a hospital, telling that his taxi number is 1729, a dull integer. No, Ramanujan replied quickly, 1729 is very interesting. It is the least positive integer that can be represented in two ways as sums of two cubes: 1729 = a3 + b3 = c3 + d3. Please give the 2 solutions. So everyone believes Ramanujan’s list. In 1948, in his PhD Thesis, M.Willerding gave another list containing ”ALL” 178 general quadratic forms (with cross-terms). [1;2;5;5] is not in this list. In 1993, J.Conway and his student W.Schneeberger gave a new list containing ”ALL” 204 such quadratic forms. CS is right. Why and how?

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Universal Quadratic Forms

Let f (X) ∈ Z[X] be a quadratic form. There is a unique symmetric matrix A ∈ 1

2Z such that f (X) = X TAX.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Universal Quadratic Forms

Let f (X) ∈ Z[X] be a quadratic form. There is a unique symmetric matrix A ∈ 1

2Z such that f (X) = X TAX.

x2 + 2y2 =

• x

y 1 2 x y

• . This is

integer-matrix quadratic form.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Universal Quadratic Forms

Let f (X) ∈ Z[X] be a quadratic form. There is a unique symmetric matrix A ∈ 1

2Z such that f (X) = X TAX.

x2 + 2y2 =

• x

y 1 2 x y

• . This is

integer-matrix quadratic form. 6x2 − 5xy + 2y2 =

• x

y

• 6

−2.5 −2.5 2 x y

• .

This is integral quadratic form.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Universal Quadratic Forms

Let f (X) ∈ Z[X] be a quadratic form. There is a unique symmetric matrix A ∈ 1

2Z such that f (X) = X TAX.

x2 + 2y2 =

• x

y 1 2 x y

• . This is

integer-matrix quadratic form. 6x2 − 5xy + 2y2 =

• x

y

• 6

−2.5 −2.5 2 x y

• .

This is integral quadratic form. If for all nonzero X ∈ Rn, f (X) > 0, then f and its matrix A are positive definite.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Fifteen Theorem for Universal Quadratic Forms

Fifteen Theorem. (1993, CS) Let f (X) ∈ Z[X] be a positive definite integer-matrix quadratic form. Then f is universal ⇐ ⇒ f represents ALL natural numbers ≤ 15!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Fifteen Theorem for Universal Quadratic Forms

Fifteen Theorem. (1993, CS) Let f (X) ∈ Z[X] be a positive definite integer-matrix quadratic form. Then f is universal ⇐ ⇒ f represents ALL natural numbers ≤ 15! Indeed, enough to represent 1,2,3,5,6,7,10,14 and 15.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Fifteen Theorem for Universal Quadratic Forms

Fifteen Theorem. (1993, CS) Let f (X) ∈ Z[X] be a positive definite integer-matrix quadratic form. Then f is universal ⇐ ⇒ f represents ALL natural numbers ≤ 15! Indeed, enough to represent 1,2,3,5,6,7,10,14 and 15. [1;2;5;5] can’t represent 15.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

CS’ Idea: Algebra+Geometry=Everything (❆ Û+➇ê➜❳➃✠Ø✹)

Local-Global Principle is valid for quadratic forms. A theorem holds in Q ⇐ ⇒ it holds in R and every p-adic ring Zp = {x ∈ R | x = ∞

n=0 anpn, an = 0, 1, · · · , p − 1}

Metric: |x| = p−m, m is the first nonzero coefficient. So lim

n→∞ pn = 0,

2+2×3+2×32 +· · ·+2×3n +· · · = −1!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

CS’ Idea: Algebra+Geometry=Everything (❆ Û+➇ê➜❳➃✠Ø✹)

Local-Global Principle is valid for quadratic forms. A theorem holds in Q ⇐ ⇒ it holds in R and every p-adic ring Zp = {x ∈ R | x = ∞

n=0 anpn, an = 0, 1, · · · , p − 1}

Metric: |x| = p−m, m is the first nonzero coefficient. So lim

n→∞ pn = 0,

2+2×3+2×32 +· · ·+2×3n +· · · = −1! Quadratic forms↔Symmetric matrices↔Lattices

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

CS’ Idea: Algebra+Geometry=Everything (❆ Û+➇ê➜❳➃✠Ø✹)

Local-Global Principle is valid for quadratic forms. A theorem holds in Q ⇐ ⇒ it holds in R and every p-adic ring Zp = {x ∈ R | x = ∞

n=0 anpn, an = 0, 1, · · · , p − 1}

Metric: |x| = p−m, m is the first nonzero coefficient. So lim

n→∞ pn = 0,

2+2×3+2×32 +· · ·+2×3n +· · · = −1! Quadratic forms↔Symmetric matrices↔Lattices L ∈ Rn is an n-dimensional lattice ⇐ ⇒ L = Zv1 ⊕ Zv2 ⊕ · · · ⊕ Zvn.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

CS’ Idea: Algebra+Geometry=Everything (❆ Û+➇ê➜❳➃✠Ø✹)

Local-Global Principle is valid for quadratic forms. A theorem holds in Q ⇐ ⇒ it holds in R and every p-adic ring Zp = {x ∈ R | x = ∞

n=0 anpn, an = 0, 1, · · · , p − 1}

Metric: |x| = p−m, m is the first nonzero coefficient. So lim

n→∞ pn = 0,

2+2×3+2×32 +· · ·+2×3n +· · · = −1! Quadratic forms↔Symmetric matrices↔Lattices L ∈ Rn is an n-dimensional lattice ⇐ ⇒ L = Zv1 ⊕ Zv2 ⊕ · · · ⊕ Zvn. A positive quadratic form is the Norm(=square of the length) of a lattice.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem

x2 − xy + y2 =

• x

y

• 1

−0.5 −0.5 1 x y

• is the

norm of the following lattice under basis v1 = (1, 0), v2 = (−1

2, √ 3 2 )

✲ ✲ ❏ ❏ ❪

v1 v2

✻

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 1)

If a quadratic form represents all positive integers, it represents 1. So CS starts their proof from the unique 1-dimensional quadratic form x2, which is the lattice Z of the real line R.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 1)

If a quadratic form represents all positive integers, it represents 1. So CS starts their proof from the unique 1-dimensional quadratic form x2, which is the lattice Z of the real line R. Now add any vector α to make a 2-dimensional lattice of the plane R2 with matrix 1 a a b

• (what’s the quadratic

form?).

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 1)

If a quadratic form represents all positive integers, it represents 1. So CS starts their proof from the unique 1-dimensional quadratic form x2, which is the lattice Z of the real line R. Now add any vector α to make a 2-dimensional lattice of the plane R2 with matrix 1 a a b

• (what’s the quadratic

form?). Since 2 must be a norm, only two possibilities occur: 1 1

• ,

1 2

• . Since the lattice of

1 1 1 2

• is

Z2, the same as 1 1

• .

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 2)

From these two 2-dimensional lattices, only the following nonisomorphic 3-dimensional lattices (totally 9) of R3 can

• ccur.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 2)

From these two 2-dimensional lattices, only the following nonisomorphic 3-dimensional lattices (totally 9) of R3 can

• ccur.

x2 + y2 + z2, x2 + y2 + 2z2, x2 + y2 + 3z2 x2 + 2y2 + 2z2, x2 + 2y2 + 3z2 and

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 2)

From these two 2-dimensional lattices, only the following nonisomorphic 3-dimensional lattices (totally 9) of R3 can

• ccur.

x2 + y2 + z2, x2 + y2 + 2z2, x2 + y2 + 3z2 x2 + 2y2 + 2z2, x2 + 2y2 + 3z2 and   1 2 1 1 3  ,   1 2 1 1 4  ,   1 2 1 1 5  ,   1 2 5  .

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 3)

From these 9 nonisomorphic 3-dimensional lattices, only 207 nonisomorphic 4-dimensional lattices of R4 can occur.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 3)

From these 9 nonisomorphic 3-dimensional lattices, only 207 nonisomorphic 4-dimensional lattices of R4 can occur. Key Lemma. If an integer-matrix form represents every positive integer up to c (to be found!) then it is universal.–Proof uses Local-Global Principle.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 3)

From these 9 nonisomorphic 3-dimensional lattices, only 207 nonisomorphic 4-dimensional lattices of R4 can occur. Key Lemma. If an integer-matrix form represents every positive integer up to c (to be found!) then it is universal.–Proof uses Local-Global Principle. There are 3 lattices missing one or more of 9 critical numbers: 1, 2, 3, 5, 6, 7, 10, 14, 15. The other 204 have the same c = 15! So totally 204 universal quaternary quadratic forms!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 3)

From these 9 nonisomorphic 3-dimensional lattices, only 207 nonisomorphic 4-dimensional lattices of R4 can occur. Key Lemma. If an integer-matrix form represents every positive integer up to c (to be found!) then it is universal.–Proof uses Local-Global Principle. There are 3 lattices missing one or more of 9 critical numbers: 1, 2, 3, 5, 6, 7, 10, 14, 15. The other 204 have the same c = 15! So totally 204 universal quaternary quadratic forms! The following 9 forms miss exactly one number.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Proof of Fifteen Theorem (continued 3)

From these 9 nonisomorphic 3-dimensional lattices, only 207 nonisomorphic 4-dimensional lattices of R4 can occur. Key Lemma. If an integer-matrix form represents every positive integer up to c (to be found!) then it is universal.–Proof uses Local-Global Principle. There are 3 lattices missing one or more of 9 critical numbers: 1, 2, 3, 5, 6, 7, 10, 14, 15. The other 204 have the same c = 15! So totally 204 universal quaternary quadratic forms! The following 9 forms miss exactly one number. [2,2,3,4]-1, [1,3,3,5]-2, [1,1,4,6]-3, [1,2,6,6]-5, [1,1,3,7]-6, [1,1,1,9]-7, [1,2,3,11]-10, [1,1,2,15]-14, [1,2,5,5]-15

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Why 15 Is Really Special?

If a quadratic form represents all positive integers below 15, then it represents all positive integers above 15!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Why 15 Is Really Special?

If a quadratic form represents all positive integers below 15, then it represents all positive integers above 15! There are forms which miss infinitely many integers starting from any of the eight critical numbers not equal to 15.

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Why 15 Is Really Special?

If a quadratic form represents all positive integers below 15, then it represents all positive integers above 15! There are forms which miss infinitely many integers starting from any of the eight critical numbers not equal to 15. 290 Theorem. (1993, Conway - 2005, M.Bhargava + J.Hanke) If a quadratic form represents all positive integers up to 290, then it represents all positive integers!

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Conway’s Soldier: A Game Created in 1961

Game Goal: Jumping as Higher as Possible on the infinite chessboard

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Conway’s Soldier: A Game Created in 1961

Game Goal: Jumping as Higher as Possible on the infinite chessboard Game Rule: Jumping and killing

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Conway’s Soldier: A Game Created in 1961

Game Goal: Jumping as Higher as Possible on the infinite chessboard Game Rule: Jumping and killing Need at least 2, 4, 8 respectively to arrive at Level 1, 2, 3

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Conway’s Soldier: A Game Created in 1961

Game Goal: Jumping as Higher as Possible on the infinite chessboard Game Rule: Jumping and killing Need at least 2, 4, 8 respectively to arrive at Level 1, 2, 3

• Question. To send a soldier to Level 4, how to arrange the

soldiers?

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Play Conway’s Soldier

• Answer. To send a soldier to Level 4, at least 20 (not

24 = 16) soldiers, arranged as the following:

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Conway and His Mathematics

Last Question. To send a soldier to Level 5, how to arrange the soldiers?

Figure: Conway&His Son

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Many Thanks but What’s Mathematics?

Mathematics is what lost in the real world and alive only in pure souls.(ê➷✸➍➢➙❦✢, ✸➚✳➙❬✮.)

Magic 15 Background Two Square Theorem Galois Finite Fields Proofs of 2-Square Theorem Three Square Theorem Four Square Theorem Universal Polynomials Magic 15 Proof of Fifteen Theorem 15 and 290

Many Thanks but What’s Mathematics?

Mathematics is what lost in the real world and alive only in pure souls.(ê➷✸➍➢➙❦✢, ✸➚✳➙❬✮.) Thank you!