The proposition is: for any integer which has the form of $n^2+1$, if we factor it, it has prime factor $p=4k+1$ for some integer $k$.
how can I proof it?
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If you are familiar with the order of some integer $n>1$ modulo some number $p$ or with quadratic residues, there is an easy proof. Not sure whether there is an even easier proof avoiding those concepts. – Peter Nov 08 '21 at 13:29
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The possible prime factors of $n^2+1$ are $2$ and the prime numbers of the form $4k+1$. To show that there must be a prime factor of the form $4k+1$ in the case $n>1$ , refute $n^2+1=2^s$ for some integer $s>1$ by considering the congruence modulo $4$. – Peter Nov 08 '21 at 13:32
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I seem to have asked something similar some time ago. https://math.stackexchange.com/questions/1676319/if-a-prime-p2-is-expressible-as-p-a2b2-then-4-mid-p-1 – AlvinL Nov 08 '21 at 13:42
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refute n^2+1=2^s for some integer s>1.. and then? – hwiba12 Nov 08 '21 at 14:39