Special Pythagorean triplet

We have a Pythagorean triplet: 8, 15, 17 where $8+15+17=40$ .

We know that if $a^2+b^2=c^2$ , then $(ak)^2+(bk)^2=(ck)^2$

Since $40 \times 25=1000$ , then $(8 \times 25)^2+(15 \times 25)^2=(17 \times 25)^2$ and $(8+15+17) \times 25 = 1000$

So $abc=8 \times 15 \times 17 \times 25^3=31875000$

$(a+b+c)^2=1000^2 \Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=2^6 *5^6 \Rightarrow c^2+ab+ac+bc=2^5 *5^6$

$\Rightarrow (c+b)(a+c)=2^5 *5^6$

The only factoring possible is $a+c=625, b+c=800$ $\Rightarrow b-a=175 \Rightarrow a^2 + (a+175)^2=(1000-175-2a)^2 \Rightarrow a=200, b=375, c=425$

$\Rightarrow abc=31875000$