$\pmod{2}$ is trivial, so we need only consider $\pmod{5}$. Essentially we want to guarantee the existance of $x,y\text{ s.t. }\forall 2\nmid m,n,\ x^{m-n}\equiv y^{m-n}\pmod{5},\text{ or }5\mid xy$. Suppose then that all of $x,y,z$ are nonzero $\pmod{5}$, and let $2\mid k$ be such that $k=m-n$. We have $xy^{-1}\equiv \pm 1\pmod{5}$. By partitioning $\left(\mathbb Z/5\mathbb Z\right)^{\times}$ into $\{1,-1\},\{2,-2\}$ the result follows from the PHP.

Let $m = 2s + 1, n = 2t + 1$, then the condition becomes $$10 \mid xy (x^{2s} y^{2t} - x^{2t} y^{2s}) = K.$$This is clearly divisible by $2$. If any of $a, b, c$ are divisible by $5$, then let $x$ be that number divisible by $5$, and we get $5 \mid K$. Otherwise, choose $x, y$ such that $x^2 \equiv y^2 \pmod 5$ (this is allowed by Pigeonhole Principle since only $1, 4$ are quadratic residues $\pmod 5$) and $5 \mid K$ in this case as well. Hence $10 \mid K$ in any case.