# Divisibility of fourths by seven

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!

## 1 Answer

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, $$a^4+b^4+c^4 \equiv 1^2+2^2+4^2\pmod 7$$.

• Is that a special, well-known rule that modulo 7 of squares is 0, 1, 2 and 4? – Pygmalion Mar 31 at 7:38
• @Pygmalion I'd say it's well-known, but nothing special. – Lord Shark the Unknown Mar 31 at 7:48
• @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. – badjohn Mar 31 at 8:04
• @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. – Pygmalion Mar 31 at 8:17
• @Pygmalion In a problem like this, with a small modulus, I would start by just writing out a table of squares. – badjohn Mar 31 at 8:19