“n = 2 is great, but the fun starts at n = 3”- Math Jorge Torres

Fermat and Euler Fermat’s Last Conjecture: There are no nonzero positive integer solutions x, y and z for the equation xn+yn=zn for n>2. Euler “kinda” proved Fermat’s last conjecture for when n = 3. Euler’s proof is a proof by contradiction that utilizes infinite descent.

Important Principles Well-Ordering Principle: Every non-empty set of positive integer has a least element. Contradiction by Infinite Descent: “Let P be a property that integers may or may not possess. If an assumption that a positive integer n has property P, which leads to the existence of a smaller positive integer N<n that also satisfies P, then no positive integer has that property.”

Euler’s Mistake Euler claimed that for to be a cube, it is suffice to find integers a and b such that p and q are given by . More so, that if is a cube, then it’s factors , because they are relatively prime, are themselves cubes….although he provided no proof. There is a major gap that confuses necessary and sufficient conditions.

The Amending Lemma Bridges the gap in Euler’s proof by utilizing Euler’s own work in sums of squares. Demonstrates characteristic concepts for numbers of the form . The Lemma states:

Building the Amending Lemma

Amending Lemma Let If this factorization contains k factors of 4, then (22)k is the largest power of which divides a2+3b2 Now, because a2+3b2 is a cube, then k is a multiple of 3 Moreover, any odd prime P must be iterated as a multiple of 3 in the factorization. That is, each of the factors is repeated 3 times so they are arranged in sequential order.

Building Lemma 2

…

Amending Lemma given by lemma 8, the factors corresponding to each group of three P’s are identical . It follows that in the factorization , , by lemma 9, then the only choice is the choice of sign and both signs cannot occur because a and b are relatively prime. then taking one factor from the group of three’s for each k and multiplying them together gives a number such that Since, The desired conclusion follows.

Preliminary The sum is odd if and only if the two addends have opposite parity. If AB = C3 and the greatest common divisor (A, B) = 1, then A and B are cubes. For the Diophantine equation xn + yn = zn , if there exist a common divisor that divides two of the three variables, then by virtue it divides the third.

Euler’s proof when n = 3 Assume towards contradiction there exist non-zero integer solutions x, y and z for x3+y3 = z3. Now, assume that x, y and z are pairwise relatively prime, and if a factor divides 2 of the 3 variables then by virtue it divides the 3rd. Pairwise relative prime implies that two are odd and one is even. Let the odds be on the left side of the equal sign. And the even on the right of the equal sign. Since the set of solutions for the Diophantine is non-empty, by the Well-Ordering Principle we can assume there exist a solution that has the smallest positive even integer.

… Suppose x and y are odd. Then z is even. x + y = 2p x – y = 2q where p and q are positive integers. p and q are of opposite parity.

Making sure p and q are Positive p is negative, only if x + y is negative. Therefore q is negative only if x – y is negative. Therefore switch x – y to y – x. And

The Diophantine Equation in p & q The Diophantine equation in terms of p and q. 2p and (p2+3q2) have opposite parity.

A Fork In the Road This proof splits into two cases, the first in which 2p and (p2+3q2) are relatively prime and both cubes, and the second case in which they are not.

Case I Assume 3 does not divide p. Since p and q are relatively prime and of the form (p2+3q2) , then by the Amending lemma there exist an a and b such that

… Then So: Expanded: (a, b) = 1

… adf If you multiply p by 2, then: And

…

Case II Assume 3 divides p and (p, q) = 1, then Now, 3 cannot divide q because (p, q ) =1, so (s, q) = 1. Then

… (a, b) = 1. For take (32• 2s) and replace s with:

… It follows that,

… Now recall that,

Conclusion Euler had a great proof, but he based it on suffice conditions, opposed to necessary conditions, and on the assumption that numbers of the form exhibit unique factorization without proving it. We saw that using Euler’s own work on sum’s of squares it is possible to construct an amending lemma to fortify Euler’s proof.

Thank you Thank you to Professor Tollisen, Professor Daigle and the Math department. Thank you to all my friends for the support.

Questions?