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. 
 A: 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. 
A: 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
$$N=0,101,-2500,2601$$
A: so  for discriminant you get
$10201+4*a^2$
first of all from the begining it is clear that this  has at least two value of $N$,namely $N=0$ and  $N=101$,now about discriminant,  what could be value of $a$,so that sum of this is perfect square?
A: The discriminant is $4a^2+101^2$ and you want this to be a square $b^2$ so we want to solve 
$(b-2a)(b+2a)=b^2-4a^2=101^2$. So since 101 is prime we need to solve $b-2a=b+2a=101$; $b-2a=1, b+2a=101^2$; $b-2a=-101, $b+2a=-101$; $b-2a=-1, b+2a=-101^2$. 
So the solutions for $N$ are now easy to find.
