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At first, I tried to proof by contradiction: I considered two numbers that are not divisible by $3$. Than I tried to write consecutive perfect squares that are divisible by $3$ to see some pastern, but it didn't get me anywhere. Then i tried to make express each whole number a,b and c under the condition that $a^2+b^2=c^2$ in terms of two arbitrary integers. I found out that $a$ has to be of the form $2mn$ and $b$ has to be of the form $m^2-n^2$, where $m$ and $n$ are whole numbers. So that I have to prove that either $2mn$ or $m^2-n^2$ is divisible by $3$. I got stuck at this point and don't know how to prove that, so help me please

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    $\begingroup$ $$x^2\equiv0\text{ or }1\mod{3}$$ $\endgroup$ Feb 1 '20 at 22:37
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HINT. Write $c^2=a^2+b^2$ and consider the equation $\mod 3$, remembering that $x \equiv 0 \mod 3$ implies that $x$ is divisible by $3$.

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Squares are $0$ or $1$ modulo $3$, so a sum of two of the latter can't be a square.

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Your first strategy will turn out to be the more effective one.

Obviously, if a number is divisible by 3, then its square will be as well. Otherwise, the number is of the form $3k\pm1$ for some integer $k$, and $$(3k\pm1)^2=9k^2\pm6k+1\equiv1\pmod3$$

Therefore, the square of a number is either divisible by 3 or it leaves a remainder of 1.


Now, let us consider that there are positive integers $a,b,c$ such that $a^2+b^2=c^2$. From our earlier notion, it cannot be that neither $a$ nor $b$ are multiples of 3, because then their sum would be of the form $3k+2$ and could not be a perfect square. Therefore, either $a$ or $b$ (or both) are divisible by 3.

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  • $\begingroup$ Thanks for the answer, but i have two questions. Why the number of the form 3k+2 can't be a perfect square? And if only one of these numbers is divisible by 3 then the sum of their squares will be of the form 3k+1? $\endgroup$
    – Ostap
    Feb 1 '20 at 23:31
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    $\begingroup$ Any number is $3k$ or $3k\pm1$, and none of those square to $3n+2$ $\endgroup$ Feb 2 '20 at 0:23

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