anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Mr.Math
  • Mr.Math
I have to go now, I will come back to it in a bit.
anonymous
  • anonymous
Take Values And Substitute!!
anonymous
  • anonymous
I suppose \( x,y \in \mathbb{N} \)

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amistre64
  • amistre64
i remember there being a trig equation to test for primes ....
anonymous
  • anonymous
Really amsitre? I never heard of it...
amistre64
  • amistre64
its in one of the boks upstairs about primes i believe
Mr.Math
  • 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.

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