Determine all triples $(x, y, z)$ of integers that satisfy the equation $x^2+ y^2+ z^2 - xy - yz - zx = 3$
Problem
Source: Bundeswettbewerb Mathematik 2023 Round 1 P2
Tags: bundeswettbewerb, algebra, math competition
iniffur
10.03.2023 03:34
Sol.
$\Longrightarrow 2x^2+2y^2+2z^2-2xy-2xz-2yz= (x-y)^2+(x-z)^2+(y-z)^2=6=1+1+4$
$\Longrightarrow (x,y,z)=(t,t+1,t+2), (t,t-1,t-2),~~~ t\in Z_0$ and permutations thereof.
TheEpicCarrot7
10.03.2023 04:04
$x^2+y^2+z^2-xy-yz-zx=(y-x)^2+(z-y)(z-x)=3$ and assume $x\leq y \leq z$. Then we realize that either $y-x=0$ and $z-y$ and $z-x$ must be $1$ and $3$ in some order which is impossible or $y-x=1$ and $z-y$ and $z-x$ must be $1$ and $2$ in some order which works but only if $z-y=y-x=1$, so the $3$ numbers must be $m-1$ $m$ and $m+1$ for some integer $m$. Plugging this in we realize all $m$ work and done.
re-express the equation as $(x+y+z)^2-3(xy+yz+zx)=3$. Clearly $x+y+z$ is a multiple of $3$ so $(x+y+z)^2$ is a multiple of $9$. This means that $xy+yz+zx$ has to be $2$ in modular $3$. Doing casework we realize that ${x mod 3, y mod 3, z mod 3}={0,1,2}$. Not sure where to go with this though.
lifeismathematics
10.04.2023 19:02
we multiply with two and observe that it comes down as $(x-y)^2+(y-z)^2+(z-x)^2=6$ which gives us $x-y=2 ,x-z=1$ and $y-z=-1$ and $x-y=-2,x-z=-1$ and $y-z=-1$ which gives $(x,y,z) \in \boxed{\{(k,k+1,k+2),(k-2,k-1,k)\} , k \in \mathbb{Z}}$ and it's permutations $\blacksquare$
mathmax12
22.05.2023 01:10
so we simplify to get (y-z)^2+(z-x)^2+(x-y)^2=6 so we solve this and it is easy