Let $x, y, z$ be positive integers, and $7 \nmid xyz$. If $7^3|x^7+y^7+z^7$, show that $7^2|x+y+z$.

by Fermat's little theorem, $x^7 \equiv x \pmod7$, then $x^7+y^7+z^7\equiv x+y+z \equiv 0 $ (mod 7)

so we have $7 | (x+y+z)$. what should I do next?

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    $\begingroup$ When wanting to show $p^2\mid a$, it usually isn't straight-forward how you would use the fact that $p\mid a$ (the obvious counterexample being when $\frac ap$ has some nice form, which is not the case here). Some times it's even easier to forget $p\mid a$ and try to show $p^2\mid a$ directly. I don't know whether that's the case here, but you could consider it. $\endgroup$ – Arthur Jul 17 '17 at 12:28
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    $\begingroup$ Actually, $7^3 \mid x^7+y^7+z^7$ iff $7 \mid x,y,z$. $\endgroup$ – lhf Jul 17 '17 at 12:41

Since $$x^7+y^7+x^7=\sum_{cyc}(x^7-x)+x+y+z,$$ we see that $x+y+z$ is divided by $7$.

In another hand, $$x^7+y^7+z^7=(x+y+z)^7-7(xy+xz+yz)(x+y+z)^5+7xyz(x+y+z)^4+$$ $$+14(xy+xz+yz)^2(x+y+z)^3-21xyz(xy+xz+yz)(x+y+z)^2-$$ $$-7(xy+xz+yz)^3(x+y+z)+7x^2y^2z^2(x+y+z)+7xyz(xy+xz+yz)^2,$$ which says (see the last term) that $(xy+xz+yz)^2$ is divided by $7$ or $xy+xz+yz$ is divided by $7$.

Thus, $7x^2y^2z^2(x+y+z)$ is divided by $7^3$ and we are done!

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  • $\begingroup$ +1 Nice answer! But how were you able to write $x^7+y^7+z^7$ in terms of the elementary symmetric polynomials? $\endgroup$ – Ovi Jul 17 '17 at 14:18
  • $\begingroup$ @Ovi I like to prove inequalities and I just know this writing. $\endgroup$ – Michael Rozenberg Jul 17 '17 at 14:27
  • $\begingroup$ My edit was to change "divided by" to "is divide by".................+1 $\endgroup$ – DanielWainfleet Jul 18 '17 at 5:16
  • $\begingroup$ @DanielWainfleet Thank you! $\endgroup$ – Michael Rozenberg Jul 18 '17 at 5:17


If $x+y\equiv0\pmod7,$ we are done


We can write $z=7a-x-y$




As $7\nmid(x+y),$ $$x+y\equiv-49a\pmod{7^3}$$

which is impossible as $7\nmid(x+y)$

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