Refreshing problem

There isn’t too much going in this question.Nonetheless,I just liked it for some bloody reason.

  • Let a and n be positive integers where a>1.If a^{n}+1 is prime,prove that a is even and n is of the form 2^{m} where m \in \mathbb{N}

Lemma:Factorization of x^{n}+1

x^{m}+1=(x+1)(x^{m-1}-x^{m-2}+x^{m-3}-\cdots+x^{2}-x+1) where m is an odd positive integer.

Simply use the factorization formula to prove this.

Proof

Since a>1,it is quite clear that a^{n}+1 \neq 2,this means that a^{n}+1 is odd and so a^{n} is even

\Rightarrow a is even.

Now,to prove the second claim,assume that n is not a power of 2.

This means that \exists some odd prime p such that p|n ie n=pk where k is a positive integer.

Here p is an odd positive integer so by the Lemma,

a^{n}+1=(a^{\frac{n}{p}})^{p}+1=(a^{\frac{n}{p}}+1)(a^{\frac{n}{p}(p-1)})-(a^{\frac{n}{p}(p-2)})+\cdots-a+1)

This even accounts for the case when n is prime ie p=n

Here the factor a^{\frac{n}{p}}+1>1,so it contradicts the fact that a^{n}+1 is prime.

This completes the proof.

 

 

 

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