Show that any set of 7 distinct integers includes two integers $x$ and $y$, such that either $x-y$ or $x+y$ is divisible by 10
Solution 1:
$$a\equiv0,\pm1,\pm2,\pm3,\pm4,5\pmod{10}\implies a^2\equiv0,1,4,9,6,5$$ So, there are $6$ in-congruent squares $\pmod{10}$
So, if we choose more than $6$ distinct integers, at least two squares will be congruent $\pmod{10}$