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 $p \equiv 1 \pmod{8}$)


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

  • $\begingroup$ 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. $\endgroup$ – confused Aug 18 '12 at 7:43
  • $\begingroup$ @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 $\equiv 1 \bmod 2*6$.) $\endgroup$ – Cocopuffs Aug 18 '12 at 8:00

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