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. 

Look at your $52$ integers $\mod 100$. So, the difference of their squares resulting in division by $100$ can be given by $a^2=b^2(\mod 100)$. This will resolute in product  of the difference of the numbers and sum of the numbers is divisible by $100$. since, any of $52$ integer numbers are asked, there can be no optimal solution answer for this.
For example:- Look at the pairs of additive inverses $(0,100)$, $(1,99)$, $(2,98)$, etc. There are $51$ such pairs. Since we have $52$ integers, two of them must belong to a pair $(a,−a)$.           Then $a^2-(−a)^2=0(\mod 100)$, so that the difference of their squares is divisible by $100$.
Likewise, since square of $10$ is $100$, so all the pair of integers with a difference of multiple of $10$ and the numbers whose additive numbers comes resulting in $\mod 10$ will also end up in the difference of squares is divisible by $100$.
For example: $(0,10)$, $(10,20)$, $(20,30)$,... etc.
Similarly, sum is multiple of $20$ and difference is multiple of $5$ will produce the same result.
so, this assumption is proved.
But the following analysis can be further improved by Chinese remainder theorem. There are $11$ distinct squares modulo $25$. By the Chinese remainder theorem, there are only $2\cdot 11=22$ distinct squares modulo $100$. So the $52$ in the problem can be improved to $23$.
since, square of $0$'s at unit place ending up in $0$, likewise $1$ in $1$, $2$ in $4$, $3$ in $9$, $4$ in $6$, $5$ in $5$, $6$ in $6$, $7$ in $9$, $8$ in $4$, $9$ in $1$. So, every square is congruent to either $0$ or $1$ or modulo $4$. so, there are only $23$ integers, for which two of the integers will draw the result that difference of their squares is divisible  by $100$.
