Only Squares Allowed

Note that the bottom can be factored as:

$z^2(x^2+y^2) = 4225$

Now let $x^2+y^2 = p, z^2 = q$ for $p,q \in \mathbb{Z}^+$

Therefore, the system now looks like:

$(x^2+y^2)+z^2 = 194 \implies p+q = 194$

$(x^2+y^2)z^2 = 4225 \implies pq = 4225$ .

Remark that $p,q$ are the roots of the quadratic: $0= t^2-194t+4225$ by vieta.

This factors as: $(t-25)(t-169) = 0$

Case 1:

$q = 25, p = 169$ $q = 25 \implies z = \pm 5$

$p = 169 \implies x^2+y^2 = 13^2$ . The trivial solutions are obviously $(x^2, y^2) = (0, 169)$ and its permutation.

Furthermore, the pythagorean triple $(5,12,13)$ also satisfies the given constraints. Finally, by FLT, this is the only non trivial combination.

From the $(0,169)$ pair, we must have one variable among $(x,y)$ be $0$ while the other can be $\pm 13$ .

This gives $4$ $(x,y)$ pairs.

From the pythagorean pair $(5,12)$ , we can have one of $(x,y)$ be $\pm 5$ while the other can be $\pm 12$

This gives us a total of $8$ $(x,y)$ pairs. Since there are two choices for $z$ , the total number of combination is $(4+8) \cdot 2 = 24$

Now note that the $(q,p) = (169, 25)$ case gives the same number of $(x,y,z)$ combinations. Therefore, the final answer is $24 \cdot 2 = \boxed{48}$

x^2 + y^2 = 194 - z^2 (x^2 + y^2)(z^2) = 4225 = (194 - z^2)(z^2) z^4 - 194z^2 + 4225 = 0 z^2 = 169 or 25 if z^2 = 169, x^2 + y^2 = 25; x^2 = 25, 16, 9 or 0; y^2 = 0, 9, 16 or 25 if z^2 = 25, x^2 + y^2 = 169; x^2 = 169, 144, 25 or 0; y^2 = 0, 25, 144 or 169

Thereafter, count the number of integer solutions from each possibility

Let x^2 = A y^2 = B z^2 = C. such that A+B+C=194 AC+BC=(A+B)C=4225 => (194-C)*C=4225 => C^2-194C+4225=0 => (C-25)(C-169)=0 Therefore C=25 or 169 => z=±5,±13

If C=25(z=>2 cases) A+B=169 = 13^2 => You can think Pythagorean theorem using only int but can use 0. The case of A+B(A,B∈int^2) is (0,13^2), (5^2,12^2), (12^2,5^2), (13^2,0). So (x,y) = (0,±13), (±5,±12), (±12,±5),(±13,0) => 12 cases! Therefore (x,y,z) = 24 cases if z^2=25

If C=169(z=> 2 cases) A+B=25 = 5^2 => You can think Pythagorean theorem using only int but can use 0. The case of A+B(A,B∈int^2) is (0,5^2), (3^2,4^2), (4^2,3^2), (5^2,0). So (x,y) = (0,±5), (±3,±4), (±4,±3),(±5,0) => 12 cases! Therefore (x,y,z) = 24 cases if z^2=169

Therefore the sum of (x,y,z) cases is 48.

(x^2+y^2)+z^2 = 194 and z^2(x^2+y^2) = 4225 solving this simultaneously gives us x^2 + y^2 = 25 or 169 and z^2 = 169 or 25

And notice that 13^2 and 5^2 are one of Phytagorean triples so that By considering all the combinations , we can get,

(x,y,z) = ( \pm12 , \pm5, \pm5 ) (\pm5,\pm12,\pm5)(\pm4,\pm3,\pm13)(\pm3,\pm4,\pm13)(\pm13,0,\pm5)(0,\pm13,\pm5)(0,\pm5,\pm13)(\pm5,0,\pm13)

Hence 48 solutions

P.S I can't use LaTeX :|

We have:

(x^2 + y^2) + z^2 =194 (x^2 + y^2) * z^2 = 4225

Solving for the two numbers we get two cases: Case 1: x^2 + y^2 = 25 and z^2 = 169 There are 4 possible tuples of non-negative answers: (0,5,13), (5,0,13), (3,4,13), (4,3,13)

Case 2: x^2 + y^2 = 169 and z^2 = 25 There are 4 possible tuples of non-negative answers: (0,13,5),(13,0,5),(12,5,5),(5,12,5)

Considering the negative cases, we get 48 answers in total

Let $a = x^2 + y^2$ and $b = z^2$ . Then $a + b = 194, ab = 4225$ . By solving the quadratic equation $M^2 - 194 M + 4225$ , we get that $(a,b) = (25,169) \mbox{ or } (169,25)$ .

When $a = 25$ , we have $x^2 +y^2 = 25$ , which gives 4 solutions of the form $(\pm 3,\pm 4)$ , 4 solutions of the form $(\pm 4,\pm3)$ , 2 solutions of the form $(\pm 5, 0)$ and 2 of the form $(0,\pm5)$ for a total of 12 solutions for $x$ and $y$ . There are 2 choices for $z$ for a total of 24 solutions.

When $a = 169$ we will have the same number of solutions again, since $169 = 13^2 + 0^2 = 5^2 + 12^2$ . So in total we have $24 + 24 = 48$ ordered triples.

x^2, y^2, z^2 > 0 => x^2, y^2, z^2 <= 194 => |x|, |y|, |z| < 20

Therefore I implemented using C++: n=20; for (int i=-n;i<=n;++i) for (int j=-n;j<=n;++j) for (int k=-n;k<=n;++k) if (i i+j j+k k==194) { int S=(i i+j j) (k*k); if (S==4225) cout << i << " " << j << " " << k << endl; }

and the result is the number of lines in output. Complexity: O(n^3) with n=20;

x^2 +y^2 +z^2 =194..........................(1). x^2 z^2 + y^2 z^2 =4225.....................(2). From the equation (1) we find that x^2 + y^2 =194-z^2........................(3). From the equation (2) we find that z^2 (x^2 + y^2) = 4225 ........................(4). Now divide the equation (4) by the equation (3), z^2 = \frac {4225}{194-z^2}. or, -z^4 + 194z^2 = 4225. or, z = (5,-5,13,-13). Now see, put the value of z in the equation (3), x^2 + y^2 = 194-25 = 169. or, x^2 +y^2 = 0^2 + 5^2 = 0^2 + (-5)^2 = 5^2 +0^2 = (-5)^2 +0^2=3^2 + 4^2 = (-3)^2 + 4^2 = 3^2 + (-4)^2 = (-3)^2 +(-4)^2 = 4^2 +3^2 = (-4)^2 + 3^2 = 4^2 +(-3)^2 = (-4)^2 + (-3)^2. So we get 12 numbers of x and y values. As there are 4 values of z, the total number of (x,y,z) is 4 \times 12 =48. And that's the answer. Ans. : 48.