anonymous 4 years ago 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.

1. Mr.Math

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

2. anonymous

Take Values And Substitute!!

3. anonymous

I suppose $$x,y \in \mathbb{N}$$

4. amistre64

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

5. anonymous

Really amsitre? I never heard of it...

6. amistre64

its in one of the boks upstairs about primes i believe

7. Mr.Math

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.