<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 want $N(N-101)$ to be a perfect square. Note that the gcd of $N$ and $N-101$ is $1$ or $101$.
Suppose it is $101$. Let $N=101M$. We get that $101M(101M-101)$ is a perfect square. This is the case iff $M(M-1)$ is a perfect square. That happens iff $4M^2-4M$ is a perfect square.
But $4M^2-4M$ is a perfect square iff $(2M-1)^2-1$ is a perfect square, say $a^2$, or equivalently iff $(2M-1)^2=a^2+1$. The only possibility is $2M-1=\pm 1$, giving $M=0$ or $M=1$, and therefore  $N=0$ or $N=101$.
Next we deal with $N$ and $N-101$ relatively prime. Then each of $N$ and $N-101$ is a perfect square, say $N=a^2$ and $N-101=b^2$. We arrive at the equation $a^2-b^2=101$. There are $4$ possibilities: $a-b=1$, $a+b=101$; $a-b=101$, $a+b=1$; $a-b=-1$, $a+b=-101$; $a-b=-101$, $a+b=-1$.
For each of these, solve for $a$ and $b$. But note that $a-b=-1$, $a+b=-101$ yields the negatives of the solutions of $a-b=1$, $a+b=101$. Thus they give the same value of $N$, as do the other pair. 
Remark: Your discriminant idea will work nicely, in basically the same way. (One could even argue that it will work more nicely). Setting $N(N-101)=a^2$, like you did, we find that the discriminant is $101^2+4a^2$. We want this to be a perfect square, say $b^2$. That gives $101^2=b^2-4a^2=(b-2a)(b+2a)$. So we want put $b-2a$ as one factor of $101^2$, and $b+2a$ as the complementary factor. We end up solving a few pairs of linear equations in two unknowns, just as in the proof given above. 
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