# Can twice a perfect square be divisible by $q^{\frac{q+1}{2}} + 1$, where $q$ is a prime with $q \equiv 1 \pmod 4$?

Can twice a perfect square be divisible by

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

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

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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 $\ldots$?". –  Jose Arnaldo Bebita Dris 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.) –  Jose Arnaldo Bebita Dris Mar 29 '13 at 6:54
@ArnieB.Dris: Never mind. I will TRY getting another solution. –  Inceptio Mar 29 '13 at 7:00

Hint:

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

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

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

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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. –  Jose Arnaldo Bebita Dris 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 :) –  Jose Arnaldo Bebita Dris Mar 30 '13 at 12:51
So it means that the condition will only work for odd perfect squares, then? :) –  Jose Arnaldo Bebita Dris Mar 30 '13 at 12:53
Theorem: Let $n$ be a positive integer. There exists a positive integer $k$ such that $n | 2k^2$