How to find three natural numbers $n_1, n_2, n_3$ with $n_1 < n_2 < n_3$ such that $n_1 + n_2 + n_3 = 784$ and $n_1^2 + n_2^2 = n_3^2$

Question

How to calculate the multiplying of the three numbers $(n_1, n_2, n_3)$ where the summation of them equals 784 if you know the followings: $n_1, n_2, n_3$ are natural numbers $n_1 < n_2 < n_3$, $n_1^2 + n_2^2 = n_3^2$

Answer 1

For our purposes we can relax the constraint $n_1<n_2$ and assume $n_1=2dab,\,n_2=d(a^2-b^2),\,n_3=d(a^2+b^2)$ for naturals $a,\,b,\,d$ with $a>b,\,(a,\,b)=1$, so $da(a+b)=392=2^3 7^2$. We need $a+b<2a$ and $(a,\,a+b)=1$, so $a+b=2^3=8,\,a=7,\,b=1,\,d=7$ and $n_1 n_2 n_3 = 2\times 7^4\times (7^4-1^4)=11524800$.

Answer 2

Hint: $7,24,25$ is a primitive Pythagorean triple. Note that $7+24+25 = 56$ divides 784.