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I am otherwise very good in mathematics, but recently I came upon a problem that I just can't solve. Do you have any idea how to solve it?

If $a^2 + b^2 + c^2$ is divisible by 7, prove that $a^4 + b^4 + c^4$ is divisible by 7 as well.

Thanks!

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Modulo $7$, squares are congruent to $0$, $1$, $2$ or $4$. The only way a trio of these can add to zero modulo $7$ is for all of them to be $0$, or for them to be some permutation of $1$, $2$ and $4$. In the latter case, $a^4+b^4+c^4 \equiv 1^2+2^2+4^2\pmod 7$.

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  • $\begingroup$ Is that a special, well-known rule that modulo 7 of squares is 0, 1, 2 and 4? $\endgroup$ – Pygmalion Mar 31 at 7:38
  • $\begingroup$ @Pygmalion I'd say it's well-known, but nothing special. $\endgroup$ – Lord Shark the Unknown Mar 31 at 7:48
  • $\begingroup$ @Pygmalion In a problem like this, it should be obvious that it is useful to know and it only takes a couple of minutes to check. I didn't happen to know that but I hope that I would have found it quickly. $\endgroup$ – badjohn Mar 31 at 8:04
  • $\begingroup$ @badjohn I suspected that there should be something special about divisibility of 7, but I only found another rule that claims that 10*a+b is divisible by 7 if a-2*b is divisible by 7. Obviously not very useful in this case. $\endgroup$ – Pygmalion Mar 31 at 8:17
  • $\begingroup$ @Pygmalion In a problem like this, with a small modulus, I would start by just writing out a table of squares. $\endgroup$ – badjohn Mar 31 at 8:19

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