\section*{Problem 6 - 320936}

a) Give three integers \(x, y\) and \(z\) such that:

\[
x^{2}+y^{2}+z^{2}-4 x+12 y-14 z-57=0
\]

b) Determine the number of all triples \((x, y, z)\) of integers \(x, y, z\) that satisfy equation (1)!
}

It is

\((x-2)^{2}+(y+6)^{2}+(z-7)^{2}=x^{2}-4 x+4+y^{2}+12 y+36+z^{2}-14 z+49=x^{2}+y^{2}+z^{2}-4 x+12 y-14 z+89\), so (1) is equivalent to

\[
(x-2)^{2}+(y+6)^{2}+(z-7)^{2}-89-57=0 \quad \text { or } \quad 146=(x-2)^{2}+(y+6)^{2}+(z-7)^{2}
\]

It is \(146=12^{2}+1^{2}+1^{2}\), so one can choose \(x-2=12, y+6=z-7=1\), i.e., \((x, y, z)=(14,-5,8)\), which provides a solution to the original equation and solves part a).

For b), we note that the number of integer solutions \((x, y, z)\) to the original equation is exactly the number of integer solutions \((a, b, c)\) to the equation \(146=a^{2}+b^{2}+c^{2}\), since from each solution of one, a unique solution of the other can be obtained via \(x-2=a, y+6=b\) and \(z-7=c\).

The number 146 can be represented as a sum of three squares in the following ways (ignoring the order of the summands):

\[
146=144+1+1=121+25+0=121+16+9=81+64+1=81+49+16
\]

For the last three representations, there are 6 possible orders of the summands and independently 2 choices for the signs of \(a, b\), and \(c\), so there are \(6 \cdot 2^{3}=48\) solutions for each; for both representations, there are 144 solutions in total.

For the second representation \(146=121+25+0\), there are again 6 possible orders of the summands, but only 2 possible signs for the non-zero squares, so there are \(6 \cdot 2^{2}=24\) solutions for this representation.

And for the first representation, there are 2 possible signs for each of the variables, but only 3 possible orders of the summands, so there are \(3 \cdot 2^{3}=24\) solutions.

In total, \(146=a^{2}+b^{2}+c^{2}\) has exactly \(144+24+24=192\) different integer solution triples, so this is also the number of solutions for the original equation (1).