Special Pythagorean triplet

We are looking for an integer sided right angle triangle with perimeter 1000. Let's check if there are any 'basic' right triangles whose perimeter is a factor of 1000.

3 + 4 + 5 = 12 :-(

5 + 12 + 13 = 30 :-(

7 + 24 + 25 = 56 :-(

8 + 15 + 17 = 40 :-) Yes! This is a factor as 40 x 25 = 1000

So scale it up 25 times to get the perimeter 1000, giving sides to be 200, 375, 425. Whose product is 31875000

Just to provide an analytic solution....

With $c = 1000 - (a + b)$ we require $0 \lt a \lt b \lt 1000$ such that

$a^{2} + b^{2} = (1000 - (a + b))^{2} \Longrightarrow a^{2} + b^{2} = 1000^{2} - 2000(a + b) + (a + b)^{2} \Longrightarrow$

$0 = 1000^{2} - 2000(a + b) + 2ab \Longrightarrow ab - 1000a - 1000b = -500000 \Longrightarrow$

$(a - 1000)(b - 1000) - 1000^{2} = -500000 \Longrightarrow (a - 1000)(b - 1000) = 500000 \Longrightarrow$

$(1000 - a)(1000 - b) = 2^{5} \times 5^{6}$ .

Now since $0 \lt a \lt b \lt 1000$ we require that $0 \lt (1000 - b) \lt (1000 - a) \lt 1000$ , so we need to look for pairs of positive integers under these restrictions whose product is $500000 = 2^{5} \times 5^{6}$ . Given this factorization, the only option is $675 \times 800$ , which leads to $(a,b,c) = (200, 375, 425)$ , and thus $abc = \boxed{31875000}$ .

Any Pythagorean triple with $a,b,c > 0$ can be generated uniquely (up to exchange of $a$ and $b$ ) as $\begin{cases} a = p^2 - q^2 \\ b = 2pq \\ c = p^2 + q^2\end{cases}\ \ \ \ \ \ \ p > q > 0.$ Adding these, we find that $a + b + c = 2p(p+q) = 1000.$ We also have that $p < p + q < 2p$ . Thus we solve for integers the system $\begin{cases} p(p+q) = 500 \\ \sqrt{500} > p > \sqrt{250},\ \ \ \text{i.e.}\ \ \ \ 22 \geq p \geq 16 \end{cases}$ It is easy to find the only solution to be $p = 20,\ q = 5,\ a = 375,\ b = 200,\ c = 425.$ (But we must flip $a,b$ to satisfy the given condition $a < b < c$ .) The answer is $200 \cdot 375 \cdot 425 = \boxed{31\,875\,000}$ .