If $x^2=10z-34$, $y^2=8x-23$, $z^2=7-6y$, find the integer value of $x+y^2+z^3$

Question

The following set of equations is given: $$\begin{align} \ & x^2=10z-34 \\ \ & y^2=8x-23 \\ \ & z^2=7-6y \end{align}\\$$ It is asked to find the integer value of $x+y^2+z^3$.

I have tried finding individual values of $x$, $y$, and $z$ but that leads to a very messy and large equation. When I try substituting values into $x+y^2+z^3$, I keep getting stuck in a loop where each variable substitution leads to the next variable and in the end I'm not able to solve it. Clearly there must be an easier way to solve this question, using simple algebra, but I'm not able to find it.

Any help leading towards the solution will be appreciated. Thank you :)

Answer 1

Here's how I'd solve it without a computer, and assuming that there's at most one solution.

Assume all variables are integers. Note that since $x^2,y^2,z^2\geq0$, $10z\geq34$, so $z\geq4$. Thus, $7-6y\geq16$, so $y\leq -2$. Thus, $8x-23\geq4$, so $x\geq4$.

So, $10z-34\geq16$, and thus $z\geq5$. So $7-6y\geq25$, and thus $y\leq-3$. Thus, $8x-23\geq9$, and so $x\geq4$.

This gives us a solution $(4,-3,5)$.

Answer 2

Hint: There is an obvious integer solution $$ (x,y,z)=(4,-3,5). $$ Substituting $6y:=7-z^2$ and $8x:=y^2+23$ the last equation gives $$ (z-5)f(x)=0 $$ with an irreducible polynomial $f\in \Bbb Z[z]$, which has no real root. Hence the above solution is the only real solution.

Answer 3

A numerical approach can be taken this way:

Set $x_0=y_0=z_0=0$

Let $x_{n+1}=\frac{{y_n}^2+23}8$

Let $y_{n+1}=\frac{7-{z_n}^2}6$

Let $z_{n+1}=\frac{{x_n}^2+34}{10}$

The values converge (but very slowly).

I had $x_{23}=3.766577037, y_{23}=-2.663788835, z_{23}=4.813051628$

You can see where it's heading at that point.