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Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.

I am not able to find an idea on how to proceed with the above questions. I have found only the obvious solution $(1,1,1)$.

Could you please provide some hints and ideas on how to proceed with the above question? Also, can we find the solutions?

Thanks.

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    $\begingroup$ another one just as obvious is $(0,0,0)$ $\endgroup$ – gt6989b Apr 7 '16 at 14:36
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    $\begingroup$ Some more obvious solutions are given by any choices of assignments $x, y, z \in \{0, 1\}$. $\endgroup$ – Dustan Levenstein Apr 7 '16 at 14:36
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    $\begingroup$ The next thing you should note is that if $x>1$, then $x^3>x^2$. So as soon as you include numbers bigger than one, you're going to need to balance them with numbers satisfying $y^3<y^2$, which, in particular, must be negative. $\endgroup$ – Dustan Levenstein Apr 7 '16 at 14:38
  • $\begingroup$ @DustanLevenstein I've realized that. We'll have to choose negative integers. $\endgroup$ – TheRandomGuy Apr 7 '16 at 14:39
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Let $z=-x$. Then $$2x^2+y^2=y^3$$ $$2x^2=(y-1)y^2$$ If $\frac{y-1}2=k^2 - $ perfect square, then $$y=2k^2+1, x=k(2k^2+1)$$ Answer: $$x=k(2k^2+1), y=2k^2+1, z=-k(2k^2+1)$$

Second method:

Let $y=1+a, z=1-a$. Then $$x^3+2(1+3a^2)=2(1+a^2)+x^2$$ $$x^2-x^3=4a^2.$$ Let $1-x=4p^2$, then $$x^2(1-x)=(4p^2-1)^24p^2=(2a)^2.$$ Let $a=p(4p^2-1)$. Then $$(x,y,z)=(1-4p^2, 1+p(4p^2-1), 1-p(4p^2-1))$$

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  • $\begingroup$ Great answer. Thanks. But how did you get the idea of putting $z=-x$? $\endgroup$ – TheRandomGuy Apr 7 '16 at 15:26
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    $\begingroup$ @Dhruv: It's always helpful to reduce the number of elements you need to work with. By setting $z=-x$, you eliminate $z$ as an independent variable, and take advantage of the fact that $x^3+(-x)^3=0$, which further reduces the number of items in the resulting equation. $\endgroup$ – Kieren MacMillan Aug 24 '16 at 20:48
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Let $y=1+a, z=1-a$. Then $$x^3+2(1+3a^2)=2(1+a^2)+x^2$$ $$x^2-x^3=4a^2.$$ Let $1-x=4p^2$, then $$x^2(1-x)=(4p^2-1)^24p^2=(2a)^2.$$ Let $a=p(4p^2-1)$. Then $$(x,y,z)=(1-4p^2, 1+p(4p^2-1), 1-p(4p^2-1))$$

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  • $\begingroup$ Where you said 'Let $1-x = 4p^2$', how did you know that $1-x$ was a multiple of 4? Is that because for integers $((2p+1)(2p-1))^2 > (2p)^2$? $\endgroup$ – TheRandomGuy Apr 7 '16 at 15:36

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