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

Can twice a perfect square be divisible by

$$q^{\frac{q+1}{2}} + 1,$$

where $q$ is a prime with $q \equiv 1 \pmod 4$?

share|cite|improve this question
Isn't that number even? – Jonas Meyer Mar 29 '13 at 6:21
Oops, sorry I missed that particular detail. I actually meant to ask: "Can twice a perfect square be divisible by $\ldots$?". – user11235813 Mar 29 '13 at 6:48
@Inceptio, can you still edit your answer? I apologize, I am unsure on how we can proceed. (I am a bit unfamiliar with how Math@StackExchange works - I only know how to do the LaTeX part.) – user11235813 Mar 29 '13 at 6:54
@ArnieB.Dris: Never mind. I will TRY getting another solution. – Inceptio Mar 29 '13 at 7:00
up vote 3 down vote accepted


$q \equiv 1 \mod 4 \implies q^{\frac{q+1}{2}} + 1 \equiv 2 \mod 4$

$q^{\frac{q+1}{2}} + 1 =(2k+1)2$

Twice of $(2k+1)^2$ is divisible by $q^{\frac{q+1}{2}}+1$. Maybe you can write $k$ in terms of $q$.

share|cite|improve this answer
Thanks for your detailed answer @Inceptio. I apologize, but I think I was in the process of editing my question (in response to JonasMeyer's comment) while you were writing down this answer. – user11235813 Mar 29 '13 at 6:53
@ArnieB.Dris: Is this a part of a question or the actual question itself? – Inceptio Mar 29 '13 at 7:28
Thanks @Inceptio, it's an actual question :) – user11235813 Mar 30 '13 at 12:51
So it means that the condition will only work for odd perfect squares, then? :) – user11235813 Mar 30 '13 at 12:53
Gladly accepting your answer now @Inceptio. – user11235813 Mar 31 '13 at 4:55

Try proving something "harder":

Theorem: Let $n$ be a positive integer. There exists a positive integer $k$ such that $n | 2k^2$


share|cite|improve this answer
Easier to Harder? ;) – Inceptio Mar 29 '13 at 7:34
Yah. Harder, in scare quotes, was what I meant. This is a more general theorem, so in principle it should be harder. But really the generalization simply brushes away all of the obfuscating detail which makes it easier to see the solution without being sidetracked. – Hurkyl Mar 29 '13 at 7:36
Thanks @Hurkyl. I do get your point. :) – user11235813 Mar 30 '13 at 12:52

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.