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32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE ictcm.com | #ICTCM

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM BY NADEEM ASLAM INSTRUCTOR OF MATHEMATICS FLORIDA INTERNATIONAL Solving UNIVERSITY MIAMI, FLORIDA Diophantine ASLAMN@FIU.EDU JAY VILLANUEVA Equations by INSTRUCTOR OF MATHEMATICS MIAMI DADE COMMUNITY COLLEGE Excel MIAMI, FLORIDA JVILLANU@MDC.EDU ICTCM, ORLANDO, FLORIDA. MARCH 12-15, 2020.

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM 1. Introduction 1.1 Brief biography of Diophantus 1.2 Framing the Diophantine Equations Solving 2. Methods of Solutions Diophantine 2.1 By Formula Equations by 2.2 By Congruence Equations 2.3 By Excel Excel 3. Examples (5) 4. Conclusions

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM 1-Introduction 1.1 Diophantine lived in Alexandria ~250 CE. Of 13 books he wrote on Arithmetica, only 6 remain, where he invented a system of notation complete with unknowns that allowed him to address Algebra problems. The only source of details about his life is an epigram found in a collection called Greek Anthology: “Diophantus passed one sixth of his life in childhood, one twelfth in youth, and one seventh as a bachelor. Five years after his marriage was born a son who died four years before his father, at half his father’s age.” How long did Diophantus live? ! ! ! ! [ 𝑦 = 84 years] 𝑦 = " + #$ + % + 5 + $ + 4 1.2 A Diophantine equation is an indeterminate equation that admits only integer solutions, of the form + 𝑕 𝑧 = ℎ(𝑨) , 𝑔 𝑦 𝑔, 𝑕, ℎ are polynomials. The linear case: 𝑏𝑦 + 𝑐𝑧 = 𝑑 , admits many interesting problems.

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM 2. Methods of Solutions 2.1 By Formula 𝑏𝑦 + 𝑐𝑧 = 𝑑 , 𝑏, 𝑐, 𝑑, 𝑦, 𝑧 are integers Let gcd 𝑏, 𝑐 = 𝑒. If d ∤ 𝑑 , then there are no solutions. If d|c, there may be unique or multiple solutions. If (𝑦 & , 𝑧 & ) a solution: 𝑏𝑦 & + b 𝑧 & = 𝑑 ⇒ 𝑏 𝑦 & + 𝑐 𝑒 𝑜 + 𝑐 𝑧 & − 𝑏 𝑒 n = 𝑑 𝑦 = 𝑦 & + 𝑐 y = 𝑧 & − 𝑏 𝑒 𝑜, 𝑒 n . Methods 2 & 3 will be explained after an example.

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Ex.1: Mr. Professor-Currency-Exchange-Problem A professor returning from Europe changes her euros and pounds into US dollars, receiving a total of $125.78. If the exchange rate is € 1 = $1.31 and £1 = $1.61. How many of euros and pounds did she exchange? Let x = number of pounds and y = number of euros . 1.61𝑦 + 1.31𝑧 = 125.78 ⟹ 161𝑦 + 131𝑧 = 12578. 𝑏 = 161, 𝑐 = 131, 𝑑 = 12578 . 𝑒|𝑑 ∴ ∃ solutions! 𝑒 = 𝑏, 𝑐 = 161 , 131 = 1: 𝑏𝑡 + 𝑐𝑢 = 𝑒; 𝑡, 𝑢 are integers. 161 𝑡 + 131𝑢 = 1 −−−−→ (1), In order to find s and t we proceed as follow; 161 = 131 O 1 + 30 1 = 3 − 2 . 1 = 3 − 8 − 3.2 . 1 = 3 . 3 − 8 . 1. 131 = 30 . 4 + 11 = 11 − 8.1 . 3 − 8 . 1 = 11 . 3 − 8 . 4 30 = 11 . 2 + 8 = 11 . 3 − 30 − 11 . 2 . 4 = 11 . 11 − 30 . 4 11 = 8 . 1 + 3 = 131 − 30 . 4 . 11 − 30 . 4 = 131 . 11 − 30 . 48. 8 = 3 . 2 + 2 = 131 . 11 − 161 − 131 . 1 . 48 3 = 2 . 1 + 1 = 131 . 59 − 161 . 48 = 161 −48 + 131(59) 2 = 1 . 2 + 0 ⟹ 𝑡 = −48 & 𝑢 = 59

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Ex. 1 (Continued) e 𝑏𝑡 + 𝑓 𝑐𝑢 = 𝑓𝑒 = 𝑑 e 1 = 12578 ⟹ e = 12578 𝑦 ! = 𝑓𝑡 = 12578 −48 = −603744 𝑧 ! = 𝑓𝑢 = 12578 59 = 742102 " $ 𝑦 = 𝑦 ! + # 𝑜, 𝑧 = 𝑧 ! − # 𝑜 %&% %'% % 𝑜 , = −603744 + = 742102 − % 𝑜 𝑦, 𝑧 ≥ 0 ⟹ 4608.73 ≤ 𝑜 ≤ 4609.33 ∴ 𝑜 = 4609 𝑦 = −603744 + 131 4609 = 𝟒𝟔 𝒒𝒑𝒗𝒐𝒆𝒕 y = 742102 − 161 4609 = 𝟔𝟒 𝒇𝒗𝒔𝒑𝒕

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Method2: Congruence equations a𝑦 + 𝑐𝑧 = 𝑑 (a) 161𝑦 + 131𝑧 = 12578 161𝑦 ≡ 12578 𝑛𝑝𝑒 131 30𝑦 ≡ 2 𝑛𝑝𝑒(131) We can try 𝑦 = 0,1,2, … , 130: ∴ 𝑦 = 35, 𝑧 = 53 Thus the answer is 𝟒𝟔 𝒒𝒑𝒗𝒐𝒆𝒕 & 𝟔𝟒 𝒇𝒗𝒔𝒑𝒕 Or (b): 131𝑧 ≡ 12578 𝑛𝑝𝑒(161) 131𝑧 ≡ 20 𝑛𝑝𝑒(161) We can try y = 0, 1, … , 160: Thus the answer is 𝟒𝟔 𝒒𝒑𝒗𝒐𝒆𝒕 & 𝟔𝟒 𝒇𝒗𝒔𝒑𝒕 ∴ 𝑧 = 53 & 𝑦 = 35.

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Method 3: By Excel Professor Currency Exchange Problem. 1.61x+1.31y=125.78 35 Number of Pounds x = 53 Number of Euros y = Objective 125.78 Constraints 78 Constraint1 96 Constraint2 0 Constraint3 0 Constraint4

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Ex.2: Dinner Problem A group dinner costs $96.00. If a lobster order costs $11.00 and a chicken order costs $8.00, how many ordered lobsters and chicken? Let 𝑦 = 𝑚𝑝𝑐𝑡𝑢𝑓𝑠 𝑝𝑠𝑒𝑓𝑠 𝑡 𝑧 = 𝑑ℎ𝑗𝑑𝑙𝑓𝑜 𝑝𝑠𝑒𝑓𝑠(𝑡) 11𝑦 + 8𝑧 = 96, 𝑏 = 11, 𝑐 = 8, 𝑑 = 96 𝑒|𝑑 ∴ ∃ solution(s)! 𝑒 = 𝑏, 𝑐 = 11,8 = 1: 𝑏𝑡 + 𝑐𝑢 = 𝑒; 𝑡, 𝑢 are integers. 11 𝑡 + 8𝑢 = 1, to find s, 11 = 8 [ 1 + 3 1 = 3 − 2 [ 1 = 3 − (8 − 3 [ 2) [ 1 8 = 3 [ 2 + 2 = 3 [ 3 − 8 [ 1 = 11 − 8 [ 1 [ 3 − 8 [ 1 3 = 2 [ 1 + 1 = 11 [ 3 − 8 [ 4 2 = 1 [ 2 + 0 = 11 3 + 8(−4) ⟹ 𝑡 = 3, 𝑢 = −4.

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Ex.2 (Continued) e 𝑏𝑡 + 𝑓 𝑐𝑢 = 𝑓𝑒 = 𝑑 e 1 = 96 ⟹ e = 96 𝑦 & = 𝑓𝑡 = 96 3 = 288 𝑧 & = 𝑓𝑢 = 96 −4 = −384 𝑦 = 𝑦 & + ' 𝑧 = 𝑧 & − ) ( 𝑜, ( 𝑜 = 288 + * = −384 − ## # 𝑜 , # 𝑜 𝑦, 𝑧 ≥ 0 ⟹ −36 ≤ 𝑜 ≤ −34.9 ∴ 𝑜 = −36 𝑝𝑠 − 35 ∴ 𝑗 𝑦 = 288 + 8 −36 = 𝟏 𝒑𝒔𝒆𝒇𝒔𝒇𝒆 𝒎𝒑𝒄𝒕𝒖𝒇𝒔. 𝑧 = −384 − 11 −36 = 𝟐𝟑 𝒑𝒔𝒆𝒇𝒔𝒇𝒆 𝒅𝒊𝒋𝒅𝒍𝒇𝒐. Check: 11(0)+8(12)=$96. ∴ 𝑗𝑗 𝑦 = 288 + 8 −35 = 𝟗 𝒑𝒔𝒆𝒇𝒔𝒇𝒆 𝒎𝒑𝒄𝒕𝒖𝒇𝒔. 𝑧 = −384 − 11 −35 = 𝟐 𝒑𝒔𝒆𝒇𝒔𝒇𝒆 𝒅𝒊𝒋𝒅𝒍𝒇𝒐. Check: 11(8)+8(1)=$96.

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Method2: Congruence equations a𝑦 + 𝑐𝑧 = 𝑑 𝑏 11𝑦 + 8𝑧 = 96 11𝑦 ≡ 96 𝑛𝑝𝑒 8 3𝑦 ≡ 0 𝑛𝑝𝑒(8) We can try 𝑦 = 0, 1, … , 7: & ∴ 𝑗 𝑦 = 0, 𝑧 = 12 ii 𝑦 = 8, 𝑧 = 1. Or (b): 8𝑧 ≡ 96 𝑛𝑝𝑒(11) 8𝑧 ≡ 8 𝑛𝑝𝑒(11) We can try y = 0, 1, … , 10: & 𝑗𝑗 𝑦 = 0, 𝑧 = 12 . ∴ 𝑗 𝑧 = 1, 𝑦 = 8

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Method 3: By Excel Dinner Problem: 11x + 8y = 96 First Possible Solution: Second Possible Solution: 0 8 Number of Lobster Ordered x = Number of Lobster Ordered 12 1 Number of Chicken Ordered y = Number of Chicken Ordered 96 96 Objective Objective Constraints: 0 8 Constraint1 Constraint1 12 1 Constraint2 Constraint2 0 0.1 Constraint3 Constraint3 0 0 Constraint4 Constraint4

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Ex.3: Pigs-Chicken-Problem In the farm the number of heads and legs of chickens and pigs add to 70. How many chickens and pigs were there? Let 𝑦 = 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑑ℎ𝑗𝑑𝑙𝑓𝑜 𝑧 = 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑞𝑗𝑕𝑡 5𝑦 + 3𝑧 = 70 , 𝑏 = 5, 𝑐 = 3, 𝑑 = 70 𝑒|𝑑 ∴ ∃ solution(s)! 𝑒 = 𝑏, 𝑐 = 5,3 = 1: 𝑏𝑡 + 𝑐𝑢 = 𝑒; 𝑡, 𝑢 are integers. 5𝑡 + 3𝑢 = 1, to find s, 1 = 3 − 2 K 1 = 3 − ( 5 = 3 K 1 + 2 5 − ) 3.1 . 1 3 = 2.1 + 1 = 3.2 − 5.1 = 5 −1 + 3(2) ⟹ 𝑡 = −1 & 𝑢 = 2 . 2 = 1.2 + 0

32 nd International Conference on Technology in Collegiate Mathematics VIRTUAL CONFERENCE #ICTCM Ex.3 (Continued) e 𝑏𝑡 + 𝑓 𝑐𝑢 = 𝑓𝑒 = 𝑑 e 1 = 70 ⟹ e = 70 𝑦 1 = 𝑓𝑡 = 70 −1 = −70 𝑧 1 = 𝑓𝑢 = 70 2 = 140 𝑦 = 𝑦 1 + 2 𝑧 = 𝑧 1 − 4 3 𝑜, 3 𝑜 = −70 + 5 = 140 − 7 6 𝑜 , 6 𝑜 𝑦, 𝑧 ≥ 0 ⟹ 23.33 ≤ 𝑜 ≤ 28 ∴ 𝑜 = 24, 25, 26, 27, 𝑝𝑠 28