For how many integers $x$ in the set {$1$, $2$, $3$,..., $99$, $100$} is $x^3-x^2$ the square of an integer?

Question

For how many integers $x$ in the set {$1$, $2$, $3$,..., $99$, $100$} is $x^3-x^2$ the square of an integer?

Source: ARML 2011

$x^3-x^2$ can be factored into $x^2(x-1)$, but where would I go from there?

Answer 1

As a comment suggests, if $x-1$ is a square then $x^3-x^2$ is a square. You need to count the number of elements in your set such that $x-1$ is a square.

Conversely, if $x-1$ is not a square, then $x^3-x^2$ is not a square.

Answer 2

$x^3 - x^2 = x^2(x-1)$ is a perfect square if and only if $x - 1$ is a perfect square.

$0 < x - 1 < 98$ has (including $0$) exactly $10$ perfect squares so $x$ may equal any $x = 1,5,10, 17, .... , 82$.

And that works. If $x = n^2 +1$ then $x^3 - x^2 = x^2(x-1) = (n^2 +1)^2 *n^2 = [(n^2 + 1)n]^2$.

Likewise if $x^3 - x^2 = x^2(x-1)=N^2$ then $x-1 = \frac {N^2}{x^2}=(\frac Nx)^2$. The only rationals that are perfect squares are integers so $x$ divides into $N$. I.e. $N = x\sqrt{x-1}$. $x$ can be any number that is one more than a perfect square.