You're gonna love this (Part-2)!

$\begin{cases} x^2+y^2+z^2 = 901 & \Rightarrow x^2 + y^2 = 901-z^2 \\ x^2z^2+y^2z^2 = 152100 & \Rightarrow (901-z^2)z^2 = 152100 \\ & \Rightarrow z^4-901z^2+152100 = 0 \\ & \Rightarrow (z^2-225)(z^2-676) = 0 \\ & \Rightarrow z = \pm 15, \pm 26 \end{cases}$

$\begin{cases} z = \pm 15 & \Rightarrow x^2+y^2 & = 901-225 = 676 \\ & \Rightarrow (x,y,z) & = (0,\pm 26, \pm 15) & \Rightarrow 4 \text{ solutions} \\ & & = (\pm 26, 0, \pm 15) & \Rightarrow 4 \text{ solutions} \\ & & = (\pm 10, \pm 24, \pm 15) & \Rightarrow 8 \text{ solutions} \\ & & = (\pm 24, \pm 10, \pm 15) & \Rightarrow 8 \text{ solutions} \\ z = \pm 26 & \Rightarrow x^2+y^2 & = 901-676 = 225 \\ & \Rightarrow (x,y,z) & = (0,\pm 15, \pm 26) & \Rightarrow 4 \text{ solutions} \\ & & = (\pm 15, 0, \pm 26) & \Rightarrow 4 \text{ solutions} \\ & & = (\pm 9, \pm 12, \pm 26) & \Rightarrow 8 \text{ solutions} \\ & & = (\pm 12, \pm 9, \pm 26) & \Rightarrow 8 \text{ solutions} \\ & & \text{Total ordered } (x,y,z) & = \boxed{48}\end{cases}$

The first equation can be restated as x^2 + y^2 = 901 - z^2. Using the substitution for the second equation, we have z^4 - 901(z^2) + 152100 = 0 implying that the solutions, using the quadratic formula that z^2 = 225 or 676.

From here, we solve the two Diophantine equations:

• The main key attack here is to use the general solutions for the Pythagorean triples. Note that the solutions for x and y are interchangeable.

x^2 + y^2 = z^2 x = a^2 - b^2 (a < b is accepted) y = 2ab z = a^2 + b^2

(1) x^2 + y^2 = 225

For a^2 + b^2 = 15, y = +- 2a(sqrt(15 - a^2))

Note that x and y are both required to be integers, hence, a can either be an irrational of radical with index 2 or an integer. So we let a = sqrt(c), this time c is non-negative.

y = +-2(sqrt(15c - c^2)) for 0 <= c <= 15.

Hence, for c = 0, 3, 12, and 15, there are integral solutions of the said equation. Having (x, y, z) = (15, 0, 26), (12, 9, 26), and all permutations of x and y, as well as the sign conventions for all variables. Hence, there are 24 solutions.

(2) x^2 + y^2 = 676

Using the same process as above in (1), we obtain (x, y, z) = (10, 24, 15), (0, 26, 15), and all permutations of x and y, as well as the sign conventions for all variables. Hence. there are also 24 solutions.

Total: 48 solutions.

Already in love with it ^_^,nice problem,