2017 Rules

We have

$\begin{aligned} x^2 + y^2 + z^2 &= 2017 \quad ...(1) \\ 2x+2yz&=4034 \quad ...(2) \end{aligned}$

$(1)-(2)+1$ gives us $\begin{aligned} x^2 - 2x + 1 + y^2 - 2yz +z^2 &= -2016\\ (x-1)^2 +(y-z)^2 &=-2016 \end{aligned}$

Hence there's no solution for $(x,y,z)$ .

The system can be written as $\large \begin{cases} x^2 + y^2 + z^2 = 2017\\2x+2yz=4034\end{cases}$ Add these equations produces $x^2+2x+y^2+2yz+z^2=6051⟹(x+1)^2+(y+z)^2=6052$ . The integer 6052 can be written as the sum of two squares. So we have $(24)^2+(74)^2=6052$ and $(54)^2+(56)^2=6052$ . Now, the possible values for $x=23,53,55,73$ . By squaring each of these numbers, only $x=23$ can be the possible solution for $x$ . But comparing the possible values for $y$ and $z$ , let $z=74-y$ . Substitute $z=74-y$ to any of these equation will get $y^2-74y+1994=0$ . In order to determine what kind of roots the resulting equation holds, we test it by applying discriminant. So, the discriminant of the equation is $Δ=(74)^2-4(1)(1994)=-2500$ . Since the discriminant is negative, then $y$ and $z$ has no solutions and $x$ has no solutions also. Therefore, there are no ordered triples $(x,y,z)$ that satisfy the given system.

Guessing...