Triples

I will solve this by reducing this problem to the classic problem of solving $x^2 + y^2 = a$ . First, take the congruence modulo 3 to find that $x^2 + y^2 + z^2 \equiv 1 \mod 3$ . Remember that in modulo 3, squares are always congruent to $1$ or $0$ . So it must be that exactly one is congruent to 1 and the others are congruent to 0. But because we have perfect symmetry with the variables let's just pick $y \equiv 0 \mod 3$ , $z \equiv 0 \mod 3$ .

Let's now make the substitution $y = 3y_1, z = 3z_1$ to get the equation $x^2 + 9y_1^2 + 9z_1^2 = 2011$ . Now take the congruence modulo 9 to find $x^2 \equiv 4 \mod 9 \implies x \equiv \pm 2 \mod 9$ .

On the other hand, let's find an inequality for $x$ . The equation implies that $y^2 + z^2 = 2011 - x^2$ . This implies that $2011 - x^2 > 0 \implies 2011 > x^2 \implies 44 \geq x$ .

But remember that $x = 9k \pm 2$ . Let's first assume that $x = 9k+2$ . This means $44 \geq 9k + 2 \implies 4 \geq k$ . So we only need to check $k =0,1,2,3,4$ . In the other case $44 \geq 9k - 2 \implies 5 \geq k$ and in this case we only have to check $k=1,2,3,4,5$ . (We don't check 0 because 9*0 - 2 is negative).

This reduces the problem to 10 computations. First, you compute $2011 - x^2$ for one of the $10$ $x$ 's we have to consider. Then you quickly check if it is congruent to 3 mod 4 or has a prime power factor congruent to 3 mod 4 which would imply it is impossible to write it as a sum of two squares. But if the number can be written as a sum of two squares then you solve the equation $9y_1^2 + 9z_1^2 = 2011 - x^2$ first by dividing by 9 and then factorizing $2011 - x^2$ over the Gaussian integers.

This last part may be hard but actually, you only have to do it in 3 cases. When $x=29,7,43$ because for the other values of x, you can quickly disqualify them by checking their normal factorization and finding factors congruent to 3 mod 4. The main solutions I found after all this are:

$(29,9,33),(29,21,27),(7,21,39),(43,9,9)$ . And remember the other solutions are built by permuting these solutions so the total number of positive solutions are $3! + 3! + 3! + \binom{3}{2} = 21$ .