Sets of Perfect Squares

We don't really need to think about the set $\mathcal{T}$ ; it is only used in the problem statement to make the actual question being asked easier to state.

All numbers $x^2$ in $\mathcal{S}$ satisfy $x^2 - 16^2 \equiv 0 \pmod {1000}$ , so $(x + 16)(x - 16) \equiv 0 \pmod {1000}$ . Obviously, $x \equiv \pm 16 \pmod {1000}$ works. Are there any other solutions?

Somehow we have to fit all the factors of $1000$ into $(x+16) * (x-16)$ , which includes three $5$ s and three $2$ s. Since $x + 16$ and $x - 16$ differ by $32$ , obviously all three factors of $5$ must be in one of them. Also, if that number is not divisible by $4$ , then the second one will not be either, and not enough factors of two will be present; so one of $x \pm 16$ must have three factors of $5$ and two factors of $2$ . Thus, we see all solutions are of the form $x \equiv 0 \pm 16$ or $x \equiv 500 \pm 16 \pmod {1000}$ . Listing out possible values of $x$ :

$16, 500 \pm 16, 1000 \pm 16, 1500 \pm 16, 2000 \pm 16, 2500 \pm 16$ ...

With $x = 2500 - 16$ being the $10$ th solution. $(2500-16)^2 = 2500^2 - 180000 + 256 = 6250000 - 180000 + 256 = 6170256$ , and so the answer is $\boxed{170}$ .

It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$

Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16$ and $s-16$ are in different residue classes mod $5,$ one term must have all the factors of $5$ and thus must be a multiple of $125.$

Furthermore, each of $s+16$ and $s-16$ must have at least two factors of $2,$ since otherwise $(s+16)(s-16)$ could not possibly be divisible by $8.$

So therefore the conditions are satisfied if either $s+16$ or $s-16$ is divisible by $500,$ or equivalently if $s = 500n \pm 16.$

Counting up from $n=0$ to $n=5,$ we see that the tenth value of $s$ is $500 \cdot 5 - 16 = 2484$ and therefore the corresponding element in $\mathcal{T}$ is $\dfrac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}$