x + y + z is integer if x^2 = y + 2 and y^2 = z + 2 and z^2 = x + 2

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x + y + z is integer if x^2 = y + 2 and y^2 = z + 2 and z^2 = x + 2

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Source: 2023 NZMO - New Zealand Maths Olympiad Round 2 p5

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#1 h Sep 2, 2023, 9:38 AM • 2 Y

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Let $x, y$ and $z$ be real numbers such that: $x^2 = y + 2$ , and $y^2 = z + 2$ , and $z^2 = x + 2$ .
Prove that $x + y + z$ is an integer.

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#2 h Sep 2, 2023, 9:57 AM

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Let $p=x+y+z$ , $q=xy+yz+zx$ , $r=xyz$ . Since $(x+2)(x-2)=x^2-4=y-2$ , taking a cyclic product, $\prod_{\rm cyc}(x+2)\prod_{\rm cyc}(x-2)=\prod_{\rm cyc}(x-2)$ . If any of $x$ , $y$ , $z$ equal to $2$ , plugging back gives $x=y=z=2$ , and so $x+y+z\in\Bbb Z$ . Otherwise $1=\prod_{\rm cyc}(x+2)=4p+2q+r+8$ , so $4p+2q+r=-7$ . Similarly, $\prod_{\rm cyc}(x+1)\prod_{\rm cyc}(x-1)=\prod_{\rm cyc}(x+1)$ . If any of $x$ , $y$ , $z$ equal to $1$ , plugging back gives $x=y=z=1$ , and so $x+y+z\in\Bbb Z$ . Otherwise, $1=\prod_{\rm cyc}(x-1)=r-q+p-1$ , so $p-q+r=2$ . Directly taking a sum, $x^2+y^2+z^2=x+y+z+6$ gives $p^2-2q=p+6$ . By solving the system, $(p,q,r)=(0,-3,-1)$ or $(-1,-2,1)$ , and so $p=x+y+z\in\Bbb Z$ . We're done.

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#3 h Sep 2, 2023, 10:51 AM

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What a nice question

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#4 h Sep 14, 2023, 3:09 AM

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A trigonometric solution:

Using $\cos 2\theta=2\cos^2\theta-1$ we get $x=2\cos\theta\implies y=2\cos 2\theta\implies z=2\cos 4\theta\implies x=2\cos 8\theta$ .

Therefore $\theta=\frac{2\pi k}{8\pm 1}$ for some $k\in\mathbb{Z}$ .

This makes $(x,y,z)$ is either $(-1,-1,-1)$ or $(2,2,2)$ , or a cycle of either $\left(2\cos\frac{2\pi}{7},2\cos\frac{4\pi}{7},2\cos\frac{6\pi}{7}\right)$ or $\left(2\cos\frac{2\pi}{9},2\cos\frac{4\pi}{9},2\cos\frac{8\pi}{9}\right)$ .

The tuple $(-1,-1,-1)$ makes $x+y+z=-3$ and $(2,2,2)$ makes $x+y+z=6$ .

For the other tuples, we have $2\cos\frac{2\pi}{7}+2\cos\frac{4\pi}{7}+2\cos\frac{6\pi}{7}=-1$ because it is the sum of the primitive $7$ th roots of unity and $2\cos\frac{2\pi}{9}+2\cos\frac{4\pi}{9}+2\cos\frac{8\pi}{9}=0$ because it is the sum of the primitive $9$ th roots of unity.

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