Find all the values of $n$ such that $n(n-101) $ is a complete square.
My attempt,
$n(n-101) = t^2$ (say) This implies $(2n-101)^2=101^2+(2t)^2$ by completing square, Now i applied Fibonacci's algorithm to find Pythagorean triples to get $n=2551$ and $t=5100$. But how can I find all solutions if some more solutions exist. Kindly help.
$$\gcd(n,n-101)=\gcd(n,101)$$ Now either $\gcd(n,101)=1\lor \gcd(n,101)=101$ lets assume that $$\gcd(n,101)=1$$ Then both $n$ and $n-101$ are perfect squares so $n=t^2$ and $n-101=k^2$ so $t^2-k^2=(t-k)(t+k)=101$ since $101$ is prime $t+k=101$ and $t-k=1$ so $t=51$ and $k=50$ so one solution is $n=51^2$.
Now if $\gcd(n,101)=101$ then $n=101m$ I'll let you finish this case if you need more help feel free to tell me in comments.