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How can I prove that

$x^2 + y^2 + z^2 + 3(x+ y + z) + 5 = 0$

Has no integer solutions?

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  • 2
    $\begingroup$ Easiest is to see that if $x^2+y^2 +z^2$ is odd/even then $(x+y+z)$ is odd/even (resp.). Hence, the LHS is always an odd number. (Edit: Which is essentially Dan's answer). $\endgroup$ – M.B. Nov 26 '13 at 21:59
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Notice $a(a+3)$ is always even. This means $(x^2 + 3x) + (y^2 + 3y) + (z^2 + 3z) = -5$ is even, contradiction.

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Assume that $(x,y,z) \in \mathbb{Z}^3$ satisfies $x^2+y^2+z^2 + 3(x+y+z) + 5 = 0$. For simplicity, let $f(X,Y,Z) = X^2+Y^2+Z^2 + 3(X+Y+Z) + 5$. Then $f(x,y,z) \equiv 0 \pmod{2}$. Notice that $a^2 \equiv a \pmod{2}$ implies $x^2+y^2+z^2\equiv 3(x+y+z)\pmod{2}$, and so $f(x,y,z) \equiv 5 \equiv 1 \not\equiv 0 \pmod{2}$. Contradiction.

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Multiply through by $4$, and rewrite the resulting equation as $$(2x+3)^2+(2y+3)^2+(2z+3)^2 =7.$$

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  • $\begingroup$ Very elegant indeed. $\endgroup$ – Old John Nov 26 '13 at 22:30

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