Let $n$ be a positive integer.Show that all divisors of $4n^{2} + 1$ have the form $4k + 1$ for some
integer $k$.
Form of the divisors
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Re: Form of the divisors
It suffices to show that all prime divisors $p$ of $4n^2+1$ is of the form $4k+1$. This is obvious because $p$ must be odd and has $-1$ as a quadratic residue.
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Re: Form of the divisors
@tanmoy, I suppose you were expecting a detailed ans? Ignore this post, if that's not the case.
It can be proved if $p|x^2+y^2$ and $p\not | x,y$, then $p=4k+1$ for some $k$.
It can be proved if $p|x^2+y^2$ and $p\not | x,y$, then $p=4k+1$ for some $k$.
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