Inspired From Satvik Golechha

by completing the squares, $(n+\dfrac{25}{2})^2 -\dfrac{597}{4}=m^2\longrightarrow(2n+25)^2 -4m^2=597$ factoring, $(2n-2m+25)(2n+2m+25)=597$ now lets see all possible way to write 597 within integers $597=1\times 597=-1\times -597=3\times 199=-3 \times -199$ note that if m is a possible value of m, so is -m and hence, the sum of all m is $a=\boxed{0}$ and lets compute all n's. $\begin{cases} 2n-2m+25=1,2n+2m+25=597\longrightarrow n=137\\ 2n-2m+25=3,2n+2m+25=199\longrightarrow n=38\\ 2n-2m+25=-1,2n+2m+25=-597\longrightarrow n=-162\\ 2n-2m+25=-3,2n+2m+25=-199\longrightarrow n=-63 \end{cases}$ if you are wondering why i excluded the case $2n+2m+25<2n-2m+25$ , it is because it would give the same value of n, just the other signed value of m. so $b=137+38-162-63=-50$ hence since these were un ordered pairs, we just multiply by 2 $\therefore 2(a-b)=2(0-(-50))=\boxed{100}$