The values of $N$ for which $N(N-101)$ is a perfect square

Question

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.

Answer 1

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?

Answer 2

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.

Answer 3

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 $$N=0,101,-2500,2601$$