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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 $\ldots$?". – Arnie 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.) – Arnie Dris 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 $q$.

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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. – Arnie 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 :) – Arnie Dris Mar 30 '13 at 12:51
So it means that the condition will only work for odd perfect squares, then? :) – Arnie Dris Mar 30 '13 at 12:53
Gladly accepting your answer now @Inceptio. – Arnie Dris 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 $n | 2k^2$


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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. :) – Arnie Dris Mar 30 '13 at 12:52

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