Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let d,k be integers, with k even.

suppose d|2k

suppose d does not divide 2

suppose d does not divide k

show that d equals 2k.

(I'm really just trying to understand the 2nd last line, in this answer to the question: Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$)

share|cite|improve this question
up vote 3 down vote accepted

It's not true. Take $k=6$ and $d = 4$ for example.

share|cite|improve this answer
You're right. I see now. Could you possibly explain to me how the author of the reference concludes that d = 2k in the link I provided? I'd greatly appreciate that. – confused Aug 18 '12 at 7:43
@confused I don't know what the purpose of that answer was. It looks like he wanted to generalize the question to other powers of $k$, but this generalization isn't correct (with $k = 6$, $5$ is an odd divisor of $2^6 + 1$ and is not $\equiv 1 \bmod 2*6$.) – Cocopuffs Aug 18 '12 at 8:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.