Find all positive integers n < 200, such that $n^2$ + $(n + 1)^2$ is a perfect square
I tried the following Let $n^2$ + $(n + 1)^2 = m^2$ where $m$ is a positive integer. Now $$(m+n)(m-n) = (n+1)^2$$ I dont know how to proceed after this. I also tried expanding original equation but that was also of no help.
Hint : Note that $X^2+Y^2=Z^2$ has solutions which are Pythagorean triples and these solutions are precisely of the form $(d(u^2-v^2),d(2uv), d(u^2+v^2))$, where $u,v$ are prime to each other (possibly interchanging the X and Y coordinates). If you want only positive integer solutions you can further assume that $u>v$.
For your problem, it follows from the equality $n^2+(n+1)^2=m^2$ that $n,n+1,m$ are pairwise coprime. That means you can find $u,v$ with $u>v$ and $\gcd(u,v)=1$ such that $n=u^2-v^2,~n+1=2uv$ and $m=u^2+v^2$ ($d$ is of course 1 here).