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Show that if $a$ is a positive integer and $a^m+1$ is an odd prime, then $m=2^n$ for some nonnegative integer $n$.

I followed the hint given in the book and rewrote it like this:

Let $a^m+1=(a^k+1)(a^{k(l-1)}-a^{k(l-2)}+\cdots-a^{k}+1)$ where $m=kl$ and $l$ is odd. I understand why $a^k+1$ is greater than $1$. However, I don't understand why $a^{k(l-1)}-a^{k(l-2)}+\cdots-a^{k}+1\neq 1.$ I understand that we are trying to show it has non-trivial factorization, but I don't understand why the second part has non-trivial factorization.

Find all of the primes of the form $2^{2^n}+5$, where $n$ is a nonnegative integer.

For this question I'm stuck trying to understand why its important to discuss this in terms of $3k+1$ and $3k+2$. I am also aware the only prime in this form is $7$.

emka
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  • If the second factor is 1, then $a^m+1=a^k+1$. – Gerry Myerson Feb 01 '14 at 03:47
  • $2^{2r}\equiv1\pmod3$. – Gerry Myerson Feb 01 '14 at 03:48
  • Dead Homer provided almost a complete solution. However, you still have a reasonable doubt regarding his solution which is valid. I would explain the part 1 of your question as,$a^m+1=(a^k+1)(a^{k(l-1)}-a^{k(l-2)}+\cdots-a^k+1)$.Now,observe that the signs are alternating,so since $a$ is a natural number,and m too,so $a^{k(l-1)}>a^{k(l-2)}$ and if the number of terms in the $a$ exponent are odd, so it obviously ends with a "$+$". – Hawk Feb 01 '14 at 12:18

1 Answers1

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1) If $a^{k(l−1)}−a^{k(l−2)}+⋯−a^k+1=1$ then we would have $a^m+1 = a^k+1$. This can only happen if $a=1$ or $m=k$. But $a=1 \implies a^m+1=2$, a contradiction since 2 is not an odd prime. Therefore, $m=k$, i.e., $l=1$. So, the only odd factor of $m$ is 1. Therefore, $m$ is a power of 2.

2) $2 \equiv -1 \mod 3$. If $n>0$, then $2^n$ is even. Therefore, $2^{2^n} \equiv (-1)^{2^n} \equiv 1 \mod 3$. So, $2^{2^n} + 5 \equiv 0 \mod 3$. This shows that if $n>0$ then $2^{2^n}+5$ is divisible by 3 and so is not prime. On the other hand, if $n=0$, then $2^{2^n}+5=7$, which is prime.

Dead Homer
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