Question

Given that n is a positive integer which contains an odd factor greater than one,prove that:
x^n + y^n = p has no solutions for any prime p>2.

Answer 1

I have to go now, I will come back to it in a bit.

Answer 2

Take Values And Substitute!!

Answer 3

I suppose \( x,y \in \mathbb{N} \)

Answer 4

i remember there being a trig equation to test for primes ....

Answer 5

Really amsitre? I never heard of it...

Answer 6

its in one of the boks upstairs about primes i believe

Answer 7

You need to know that \(a^n+b^n\), where n is odd can be factorized as:
\[a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-\cdots -ab^{n-2}+b^{n-1})\]
Assuming that \(x, y \in \mathbb{N}\), let
\(x^n+y^n=(x^m)^q+(y^m)^q\), where q is an odd number. Let \(x^m=a\) and \(y^m=b\). By using the factorization above we can write
\[(a^q+b^q)=(a+b)(a^{q-1}-a^{q-1}b+a^{q-3}b^2-\cdots -ab^{q-2}+b^{q-1}).\]
For this quantity to be prime
i) \(a+b\) has to be \(1\) or
ii) \(a^{q-1}-a^{q-1}b+a^{q-3}b^2-\cdots -ab^{q-2}+b^{q-1}\) has to be \(1\).
But happens only for \(a=b=1\), which gives the prime 2. But since we're looking for \(p>2\), no such solution exists.