How to solve this algebra problem related to Pythagorean triplets?

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If $a^2+b^2=c^2 $ and $a+b+c=1000$ where a,b,c are positive integers, find the product $abc$.

First I tried this: $$(a+b)^2=(1000-c)^2$$ $$a^2+b^2+2ab=c^2+1000^2-2000c$$ $$2ab=1000^2-2000c$$

But I didn't know how to proceed from here so I tried a different method.
I tried to solve this by using the formula that generates Pythagorean triples: $m^2+1, m^2-1, 2m$ so I got: $$m^2+1 +m^2-1 +2m = 1000$$ $$2m^2+2m=1000$$ $$m^2+m=500$$

Now if I solve the quadratic equation $$m^2+m-500=0$$

then I'm getting an irrational answer. So how do I solve this?

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Well, it is not really hard to prove that $\left(\text{a},\text{b},\text{c}\right)\in\left\{\left(200,375,425\right),\left(375,200,425\right)\right\}$.