TITLE: Of any 52 integers, two can be found whose difference of squares is divisible by 100
QUESTION [47 upvotes]: 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.

REPLY [2 votes]: 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.