The Special Number 24

let $A = x^2 - 1$ , then $2A = 2(x^2 - 1) = y^2 - 1$ , $\therefore y^2 = 2x^2 - 1$

So $y$ must be odd and $y^2 = 4k + 1$ , then $x$ must be odd too.

Then it is easy to get:

• $y = 1, A = 0$

• $y = 7, A = 24$

• $y = 41, A = 840$

• $y = 239, A = 28560$

Let the number $n$ satisfies $n=a^2-1,2n=b^2-1$ , where $a, b$ are integers. Then

$a=\displaystyle \sqrt {\dfrac {b^2+1}{2}}$

So $b$ must be odd. Say $b=2m+1$ . Then $a=\sqrt {2m^2+2m+1}=\sqrt {m^2+(m+1)^2}$ .

So $m, m+1,a$ must form a Pythagorean Triplet . We know that $(3,4,5),(20,21,29),...$ are such triples.

So, next to $m=3,m=20\implies a=29,$ and $n=29^2-1=\boxed {840}$ .

Brute force:

Iterating over $x$ : let $y = 2(x^2-1)$ and test whether $(y-1)$ is a perfect square.

Yes when $x=1$ ( $x^2-1 = 0$ ), $x=5$ ( $x^2-1=24$ ) and $x=29$ ( $x^2-1=840$ ).

$841 = 29^2$ and $1681 = 41^2$ .

I assumed that the question was not looking for $0$ .