Of any 52 integers, two can be found whose difference of squares is divisible by 100 Prove that of any 52 integers, two can always be found such that the difference of their squares is divisible by 100.
I was thinking about using recurrence, but it seems like pigeonhole may also work. I don't know where to start. 

In 26 integers, by Pigeonhole Principle you have at least two whose difference is zero when divided by 25. In 52 integers you have at least 3 such integers. Pick those three integers. Again, by Pigeonhole, two of them will have the same parity. Let them be $a$ and $b$. Thus $2|(a+b)$ and $2|(a-b)$ as such $4|(a+b)(a-b)$. Since 25|(a+b)(a-b) also and 4 and 25 are relatively prime you have $100|(a^2-b^2)$.
