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How do I show that if $$m^2 + n^2 + p^2 \equiv 0 \pmod 5$$ then at least one of $\{m,n,p\}$ is divisible by 5?

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  • $\begingroup$ At least one of what is divisible by $5$? $\endgroup$ – fleablood Oct 13 '19 at 18:59
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This is quite standard. The only squares mod 5 are 0, 1 and 4. Then if you have that the sum of tree squares that gives 0, at least one of them has to be zero (just check all posibilities $1+1+1=3\neq 0,1+1+4=1\neq 0$ and so on).

Finally the only number with square 0, is zero

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If $k \equiv 0,\pm 1, \pm 2 \pmod 5$ then $k^2 \equiv 0,1,-1\pmod 5$.

So none of $m,n$ or $p$ are divisible by $5$ then none of $m,n,p \equiv 0 \pmod 5$ and none of $m^2, n^2, p^2 \equiv 0 \pmod 5$. So all of $m^2, n^2, p^2 \equiv \pm 1 \pmod 5$.

So $m^2 + n^2 + p^2 \equiv \pm 1 + \pm 1 + \pm 1$. The only possible values for those is $-3,-1,1$ or $3$. So $m^2 + n^2 +p^2$ being divisible by $5$ is impossible unless one of $m,n$ or $p$ is divisible by $5$.

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