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I need to prove this using modular arithmetic but I'm a little stuck. Could I get some pointers? Thank you.

Let $a,b$ be integers. Then $3 | a^2 + b^2$ if and only if $3 | a$ and $3 | b$

healynr
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James Anderson
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    Since a square can be only 0 or 1 mod 3 the result follows immediately. – Thomas May 18 '20 at 17:43
  • There are only nine combinations of residue classes of $a$ and $b$. You can brute force this soon enough by testing them all. Symmetries will cut it down further very fast also. – Jyrki Lahtonen May 18 '20 at 17:44
  • Suppose $3\mid a$ and $3\mid b$. Then that means that... which further means that... implying that $3\mid a^2+b^2$. Now, suppose otherwise that $3\nmid a$ or $3\nmid b$. Without loss of generality, suppose it was that $3\nmid a$. Now, it follows that $a=3k\pm 1$ in which case $a^2=3(3k^2+6k)+1$... Continuing this implies... and further implies... which implies $3\nmid a^2+b^2$ – JMoravitz May 18 '20 at 17:45
  • I also warmly recommend that you study our guide for new askers. – Jyrki Lahtonen May 18 '20 at 17:53
  • By modular arithmetic there are only three cases of what $a\mod 3$ can be and $3$ cases of what $b \mod 3$ can be so there are only $9$ cases of what $(a^2 + b^2)\mod 3$ can be. If worst comes so worst just do them all and show $(a^2+b^2)\mod 3 \equiv 0 \pmod 3 \iff a\equiv 0 \pmod 3$ and $b\equiv 0 \pmod 3$. But you can reduce this from $9$ steps to a lot fewer be making simple observations for example $x+y= y+x$ so if we test $a\mod 3\equiv x$ and $b\mod 3\equiv y$ we don't need to test $a\mod 3\equiv y$ and $b\mod 3\equiv x$. – fleablood May 18 '20 at 17:59
  • However ... a very clever and powerful thing to so note is $a\equiv 2\pmod 3\implies a^2 \equiv 1 \pmod 3$ and $a\equiv 1\pmod 3\implies a^2\equiv 1 \pmod 3$ and $a^2 \equiv 2 \pmod 3$ is NEVER possible. So there are only $4$, not $9$ options for what $(a^2 + b^2)\mod 3$ can be. You can have it be $0+0$ or $0+1$ or $1+0$ or $1+1$. – fleablood May 18 '20 at 18:05

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I'm sure you can show that $3|a, 3|b \implies 3|a^2+b^2$ fairly easily. For the other way, note that $x^2\equiv 1 \pmod 3 $ when $3$ doesn't divide $x$.

healynr
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  • Congruence notation is $a\equiv b\pmod m$. Meaning that $a-b$ is divisible by $m$. $b$ may be the remainder in integer division of $a$ by $m$, but it isn't constrained to be in the range $[0,m-1]$. OTOH the binary mod is reserved for denoting the remainder: a\bmod m= r yielding $a\bmod m =r$. – Jyrki Lahtonen May 18 '20 at 17:48
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    I think you meant "when $3$ doesn't divide $x$", not "when $x$ doesn't divide $3$" – J. W. Tanner May 18 '20 at 17:48