Mathematical Reflections 2017, Issue 5 - Problem S422

Problem:
Solve in positive integers the equation $$u^2+v^2+x^2+y^2+z^2=uv+vx-xy+yz+zu+3.$$

Proposed by Proposed by Adrian Andreescu, Dallas, USA

Solution:
The given equation can be written as $$(u-v)^2+(v-x)^2+(x+y)^2+(y-z)^2+(z-u)^2=6.$$
Since $u,v,x,y,z$ are positive integers, then $x+y \geq 2$, which gives $(x+y)^2 \geq 4$. Since $(x+y)^2 \leq 6$, we conclude that $x+y=2$, so $x=y=1$ and
$$(u-v)^2+(v-1)^2+(1-z)^2+(z-u)^2=2.$$
So, exactly two of the summands on the LHS are equal to $1$ and the other are equal to $0$. We have six cases.

(i) $(u-v)^2=(v-1)^2=1$ and $(1-z)^2=(z-u)^2=0$. From the last equations we get $u=z=1$ and from the first equations we get $v=2$.

(ii) $(u-v)^2=(1-z)^2=1$ and $(v-1)^2=(z-u)^2=0$. From the last equations we get $v=1$ and $u=z$ and from the first equations we get $u=z=2$.

(iii) $(u-v)^2=(z-u)^2=1$ and $(v-1)^2=(1-z)^2=0$. From the last equations we get $v=z=1$ and from the first equations we get $u=2$.

(iv) $(v-1)^2=(1-z)^2=1$ and $(u-v)^2=(z-u)^2=0$. From the last equations we get $u=v=z$ and from the first equations we get $u=v=z=2$.

(v) $(v-1)^2=(z-u)^2=1$ and $(u-v)^2=(1-z)^2=0$. From the last equations we get $u=v$ and $z=1$ and from the first equations we get $u=v=2$.

(vi) $(1-z)^2=(z-u)^2=1$ and $(u-v)^2=(v-1)^2=0$. From the last equations we get $u=v=1$ and from the first equations we get $z=2$.

In conclusion, $$(u,v,x,y,z) \in \{(1,2,1,1,1),(2,1,1,1,2),(2,1,1,1,1),(2,2,1,1,2),(2,2,1,1,1),(1,1,1,1,2)\}.$$