For how many values of N (positive integer) N(N-101) is a square of a positive integer?

Michael is right, but the only answer N = 2601 can be found without testing.

N(N - 101) is a perfect square exactly when 4N(N - 101) is a perfect square, say

4N(N - 101) = x² for some positive integer x. Then

4N² - 2*2*101N + 101² = x² + 101², or

(2N - 101)² = x² + 101² or let y = 2N - 101 and we'll have

y² - x² = 101² or (y - x)(y + x) = 101²

Since 101 is prime and y - x < y + x the only possibility is

y - x = 1 and y + x = 101², hence 2y = 1 + 101² = 10202,

finally y = 5101 and N = (y + 101)/2 = 2601.

Suppose n(n-101) = (n-g)^2, where n and (n-g) are positive integers. Then g in {1, …, 100} and:

n^2 - 101n = n^2 - 2gn + g^2.

So 101 = 2g - g^2/n. Thus, g^2/n must be an integer. It follows that n must be less than or equal to 100^2. So you can just test all values of n in the range {102, …, 100^2}. The only solution is n=2601.

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Edit:

@Duke: Yes, that way is nicer. =)

I'm going to ramble a bit ...

N(N-101) = N^2 - 101N

Want N(N-101) = k^2

N^2 - 101N - k^2 = 0

N = (101 + sqrt(101^2 + 4k^2)) / 2

So, N is an integer if and only if 2*sqrt(101^2 + 4k^2) is perfect square