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Prove that there do not exist integers $a, b$ and $k$ such that $a^2+b^2=4k+3$.

My approach is to assume for the purpose of contradiction that there do exist integers $a, b$ and $k$ such that $a^2+b^2=4k+3$, but I'm not sure how to do so. Any help is appreciated, thanks!

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    $\begingroup$ Hint: any square is congruent to $1$ or $0$ $\pmod{4}$. $\endgroup$ – Jordan Green Jan 30 at 3:02
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$a^2+b^2$ is odd, which says that $a^2$ and $b^2$ have an opposite parity.

Since the expression $a^2+b^2$ is symmetric, we can assume that $a$ is odd and $b$ is even.

Id est, there are integers $m$ and $n$ for which $a=2m-1$ and $b=2n,$ which gives $$(2m-1)^2+(2n)^2=4k+3$$ or $$4m^2-4m+1+4n^2=4k+3$$ or $$2m^2-2m+2n^2-2k=1,$$ which gives that $1$ is divisible by $2$, which is a contradiction.

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  • $\begingroup$ +1. Well done !!!. Nice job !!!. $\endgroup$ – Felix Marin Jan 30 at 6:14
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Hint:

Substitute:

$$a=2\lambda, b= 2\mu$$ $$a=2\lambda, b=2\mu +1$$ $$a=2\lambda+1, b=2\mu+1$$ And try expressing the results in the form $4p+q$.

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