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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.
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.

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Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2I have to go now, I will come back to it in a bit.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Take Values And Substitute!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I suppose \( x,y \in \mathbb{N} \)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0i remember there being a trig equation to test for primes ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Really amsitre? I never heard of it...

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0its in one of the boks upstairs about primes i believe

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2You need to know that \(a^n+b^n\), where n is odd can be factorized as: \[a^n+b^n=(a+b)(a^{n1}a^{n2}b+a^{n3}b^2\cdots ab^{n2}+b^{n1})\] 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 \[(a^q+b^q)=(a+b)(a^{q1}a^{q1}b+a^{q3}b^2\cdots ab^{q2}+b^{q1}).\] For this quantity to be prime i) \(a+b\) has to be \(1\) or ii) \(a^{q1}a^{q1}b+a^{q3}b^2\cdots ab^{q2}+b^{q1}\) 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.
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