Number Theory Problem

Two squares? This looks like a job for modular arithmetic! We have $m^2-n^2=k$ . $Mod 4$ isn't bad to use, since all squares have the residue of 0 or 1 when divided by 4. So, we have 3 cases

• $m^2 == 0 (mod 4) - n^2 congruent 0 (mod 4)$ resulting in $k == 0 (mod 4)$ $k$ is divisible by $4$ and there are $10000/4$ integers fitting this requirement. $2500 integers$
• $m^2 == 1 (mod 4) - n^2 == 0 (mod 4)$ resulting $k == 1 (mod 4)$ . So $k$ is $1,5,9,13$ ... or every $4n+1$ where n is a whole number. We will save this requirement for the final one
• $m^2 == 0 (mod 4) - n^2 == 1 (mod 4)$ resulting $k == 3( mod 4)$ So $k$ is $3,7,11,15$ .... or every $4n-1$ where n is a whole number. Combining what we found out from the last two cases, we conclude that $k$ can be any odd integer between $1$ and $10000$ . There are 10000/2 of them $5000 integers$

So our answer is $7500$

$m^2-n^2=(m-n)(m+n).$ $$ If m, n are consecutive integers then their difference will be $1$ and sum will be an odd integer. $Hence, m^2-n^2$ = An Odd Integer. Thus, every odd integer can be written as difference of two squares. If, $m, n$ are both odd or both even then, $2|m-n$ and $2|m+n$ => $4|(m^2-n^2)$ Every, no. divisible by 4, can also be written as difference of two squares. $5000+2500= \>7500$