# $x^7+y^7+z^7$ is divisible by $7^3$, then $x+y+z$ is divisible by $7^2$

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?

• 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. – Arthur Jul 17 '17 at 12:28
• Actually, $7^3 \mid x^7+y^7+z^7$ iff $7 \mid x,y,z$. – 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!

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

HINT:

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

Otherwise

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

$$x^7+y^7+z^7=x^7+y^7+(7a-x-y)^7$$

$$(7a-x-y)^7\equiv-(x+y)^7+7(7a)(x+y)^6\pmod{7^3}$$

$$\implies-(x+y)^7+7(7a)(x+y)^6\equiv0\pmod{p^3}$$

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

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