Pythagorean identity a^2 +b^2 = c^2 and Fermat’s last theorem; we already know t
ID: 1719919 • Letter: P
Question
Pythagorean identity a^2 +b^2 = c^2 and Fermat’s last theorem; we already know the importance of Pythagorean identity: it relates certain fundamental property of geometry to certain important relation between numbers. Indeed perpendicularity that is fundamental to architecture is equivalent with the sides of a triangle satisfying the Pythagrean identity. One may wonder what magical properties could be hidden in relations like a^3+ b^3= z^3, or any other equations such as a^n + b^n = c^n for other natural numbers n (such equations are known as Diophantine equations, and there are lots of you tube presentations on these equations, watch some of them.) The well known Fermat’s Last Theorem claims that there is no positive integer solutions to equations such as x^n + y^n = z^n for any n 3.
1. Prove that if for all prime numbers p > 2 the equation x^p + y^p = z^p has no positive integer solutions then for any integer n > 2 that is not a power of 2 the equation x^n + y^2 = z^2 has no positive integer solution.
2. Fermat proved that there are no integer solutions to the equation x^4 + y^4 = z^4 . Use this to prove that assumptions of part (d) can be relaxed, that is for all prime p > 2 the equation x^p + y^p = z^p has no positive integer solutions, and if n > 4 (which can be a power of 2) then x^n + y^n = z^n has no positive integer solutions.
Explanation / Answer
1 Let n = 2k m , where m is an odd number.
As n is not a power of 2, we deduce that m is not 1.
So m has a non-trivial odd prime factor , say.
Thus n has a non-trivial odd prime factor . Let n = pq,
if xn + yn = zn admits a positive integer solution,
then
( xq )p + (yq )p = (zq) p ,
That will contradict the assumption that the Fermat equation has no postive integer solutions for odd primes.
Thus the claim in 1) is proved.
2) The argument is entirely similar to the above.
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