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. 

As other people pointed out that even if we use $23$ instead of $52$, the result holds true,but you cannot always check out the possible remainders by squaring two digit numbers.
The problem asks for $52$ integers, to simplify the work.
Here is the proof.
List out all the numbers  ${a_1}^2,{a_2}^2,{a_3}^2,{a_4}^2 \dots \dots$ ${a_{52}}^2$ 
Now divide the numbers by $100$. Now note that dividing a number $n$ by $100$ is equivalent to extracting the last two digits of the number $n$. Now the square of any number ends with 0,1,4,5,6,9 so the possible remainder are $00,10,20,30 \dots \dots 90;01,11,21,31 \dots \dots 91;02,12,22,32 \dots \dots 92; \dots \dots ;09,19,29,39 \dots \dots 99$ However any perfect square divisible by $10$ is divisible by $100$. Hence $00 ;01,11,21,31 \dots \dots 91;02,12,22,32 \dots \dots 92; \dots \dots ;09,19,29,39 \dots \dots 99$ are the $1+10\left(5\right)$ are $51$ possible remainders.Now use Pigeonhole Principle to get two numbers which leave same remainder when divided by 100. Now subtract them.
