Given $$a^6 + b^6 + c^6 = x^6 + y^6 + z^6$$

prove that $$a^2 + b^2 + c^2 - x^2 - y^2 - z^2 \equiv 0 \bmod{9}$$

I was thinking of using $n^6 \pmod{27}$ and showing both sides have the same pattern but it's getting really confusing..

  • $\begingroup$ consider $x\equiv 0,1,2,3,4,5,6,7,8 \mod 9$ and compute $x^6\mod 9$ $\endgroup$ – Dr. Sonnhard Graubner Dec 24 '15 at 4:13
  • 5
    $\begingroup$ Note that $u^6\equiv 1,0\pmod 9$ for all $u$. So this means that the number of $\{a,b,c\}$ that are divisible by $3$ is the same as the number of $\{x,y,z\}$ that are divisible by $3$. $\endgroup$ – Thomas Andrews Dec 24 '15 at 4:14
  • $\begingroup$ @Dr.SonnhardGraubner : Instead of ^6\mod 9, try ^6\bmod 9. Then instead of $^6\mod 9$ you'll see $x^6\bmod 9$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 24 '15 at 4:15
  • 1
    $\begingroup$ It's gonna take more than modular arithmetic, because this theorem is not true of $a^6+b^6+c^6\equiv x^6+y^6+z^6\pmod{9}$. For example, $2^6+0^6+0^6\equiv 1^6+0^6+0^6\pmod{9}$. So you need something about equality. $\endgroup$ – Thomas Andrews Dec 24 '15 at 4:18
  • 1
    $\begingroup$ @Shailesh Nothing I wrote is a complete answer. I actually showed why you can't use just my first comment as the answer. $\endgroup$ – Thomas Andrews Dec 24 '15 at 4:20

so only worry about numbers between $-4$ and $4$.
Their sixth powers are $19,0,10,1,0,1,10,0,19\pmod{27}$
and their squares are $7,0,4,1,0,1,4,7\pmod{9}$
The sixth powers are either $0$ or $9A+1$; the squares are either $0$ or $3A+1$, the same $A$.


Write $$ a^6+b^6+c^6-x^6-y^6-z^6 = (a^2-x^2)(a^4+a^2x^2+x^4) + (b^2-y^2)(b^4+b^2y^2+y^4) + (c^2-z^2)(c^4+c^2z^2+z^4) ; $$ by Thomas's comment, we can also pair off in such a way that $a,x$ are either both $\equiv 0\bmod 3$ or both $\not\equiv 0\bmod 3$, and similarly for $b,y$ and $c,z$.

The non-zero squares $\bmod 9$ are $1,4,7$, and now some experimentation shows that $a^4+a^2x^2+x^4\equiv 3\bmod 9$ if $a,x\not\equiv 0\bmod 3$. Thus the expression from above becomes $$ 3(a^2-x^2+b^2-y^2+c^2-z^2) $$ when considered $\bmod 9$; note that this is also correct when the other case ($a,x\equiv 0\bmod 3$) applies. Since the difference of the sums of squares is $\equiv 0\bmod 3$ (by Thomas's comment again, and since $u^2\equiv 1\bmod 3$ for $u\not\equiv 0\bmod 3$), the claim follows now.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.