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Suppose $a, b$ and $n$ are positive integers, all greater than one. If $a^n+b^n$ is prime, what can you relate $n$ with 2?

My approach: for $a^n+b^n$ to be prime $\forall n>1$, $a$ and $b$ has to be coprimes. But how do I ascertain anything about $n?$

Ruddie
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2 Answers2

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If $n$ is odd, then $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b\pm\ldots-ab^{n-2}+b^{n-1})$.

Hagen von Eitzen
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$n$ must be of the form $n=2^k$ for some non negative integer $k$ since otherwise, you can write $n$ as $n=kl$ for some odd $k$. This implies $$ a^n+b^n=(a^l)^k+(b^l)^k=(a^l+b^l)((a^l)^{k-1}-(a^l)^{k-2}b^l\pm\ldots-(a^l)(b^l)^{n-2}+(b^l)^{n-1}) $$ which is not prime.

Jlamprong
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  • If I assume $a=3, b=5$, then $a^2+b^2=34$ which is not a prime number. Then, how can $n=2^k?$ – Ruddie May 08 '14 at 14:07
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    What I told you above is: if $a^n+b^n$ is prime then $n=2^k$ for some $k$. It doesn't mean

    if $n=2^k$ for some $k$ then $a^n+b^n$ is prime.

    Please see here

     – Jlamprong May 08 '14 at 16:31
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    I get it. Only if $n=2^k$, there is a possibility of $a^n+b^n$ to be prime. – Ruddie May 09 '14 at 11:06