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Can someone give me a hint for the following number theory problem?

Show that if $x^3 + y ^3 = z^3$, then one of $x, y, z$ must be a multiple of 7.

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  • $\begingroup$ The only solution is $x=y=z=0$ for $x,y,z$ positive integers so in a sense they are all multiples of 7. I think you are missing some information in the problem statement. $\endgroup$ – John Habert Jan 29 '14 at 1:48
  • $\begingroup$ I highly doubt that's the only solution; also in a number theory problem $x, y, z$ don't need to be positive. $\endgroup$ – Ayesha Jan 29 '14 at 1:54
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    $\begingroup$ I phrased that answer a little poorly. If you assume that $x,y,z$ are positive integers, then there are no solutions. It is a famous result. Do you have some sort of extra info like working $\mod n$ for some $n$? $\endgroup$ – John Habert Jan 29 '14 at 1:58
  • $\begingroup$ The problem doesn't assume knowledge of Fermat's Last Theorem. That being said, thank you! I didn't think of it. $\endgroup$ – Ayesha Jan 29 '14 at 1:59
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Hint: If $w$ is not a multiple of $7$, then $w^3$ is congruent to $1$ or $-1$ modulo $7$.

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Hint $\,{\rm mod}\ 7\!:\ x,y,z\not\equiv 0,\ x^3\!+y^3\!= z^3\overset{\rm square}{\Rightarrow} 2+2(xy)^3 \equiv 1\,\Rightarrow\, 2(xy)^3 \equiv -1\overset{\rm square}\Rightarrow 4\equiv 1\Rightarrow\, 7\mid 3$

Note $\ $ Generally $\,a^2,b^2,c^2\!=1,\ a+b=c\,\overset{\rm square}\Rightarrow\,2+2ab =1\,\Rightarrow\,2ab=-1\overset{\rm square}\Rightarrow\,4= 1\,\Rightarrow\, 3 = 0,\,$ hence the ring has characteristic $3$ (or is trivial).

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