<s>[INST] The values of $N$ for which $N(N-101)$ is a perfect square 
For how many values of $N$ (integer),  $N(N-101)$ is a perfect square number?

I started in this way.
Let   $N(N-101)=a^2$
  or  $N^2-101N-a^2=0.$
Now if the discriminant of this equation is a perfect square then we can solve $N$.
But I can't progress further. 
 [/INST] We have $$N^2-101N = a^2 \implies (N-101/2)^2 - (101/2)^2 = a^2 \implies (N-101/2)^2 - a^2 = (101/2)^2$$
This gives us
$$(2N-101)^2 - (2a)^2 = 101^2 \implies (2N+2a-101)(2N-2a-101)=101^2$$
Note that $101^2$ can be written as
$$101 \times 101 = 1 \times 101^2 = 101^2 \times 1 = (-101) \times (-101) = (-1) \times (-101^2) = (-101^2) \times (-1)$$
This gives us
$$a=0 \text{ and }N = 0,101$$
$$4N - 202 = 1+101^2 \implies 4N = 10202+202 \implies N = \dfrac{10404}4 = 2601$$
$$4N - 202 = -1-101^2 \implies 4N = -10202+202 \implies N = -2500$$
Hence, the possible values of $N$ are
$$N=0,101,-2500,2601$$
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