Let $x$, $y$ and $z$ be natural numbers satisfying $x^2 + y^2 + 1 = xyz$. Prove that $z = 3$.

Question

Let $x$, $y$ and $z$ be natural numbers satisfying $x^2 + y^2 + 1 = xyz$. Prove that $z = 3$.

I have managed to show that $z$ is a multiple of $3$ by looking at it modulo $3$ but not sure how to show $z=3$. Any tips?

Answer 1

One could go about this in the following way: If $x = y$, then the equation becomes

\begin{align*} 2x^2 + 1 &= x^2 z \end{align*} where the left-hand side is congruent to $1$ (mod $x$) and the right-hand side is congruent to $0$. It follows immediately that $x = 1$, and consequently that $z = 3$.

Now, suppose that $y > x$ (with no loss of generality). Then we can write $y = x+k$ for some $k \in \mathbb{Z}_+$. Substituting this into the equation, we see that

\begin{align*} x^2 + x^2 + 2kx + k^2 &= x(x+k)z \\ &\equiv 0 \end{align*} (mod $x$). Consequently,

\begin{align*} x \mid k^2 = (y-x)^2 = y^2 - 2yx + x^2 \equiv y^2 \end{align*} (mod $x$).

Thus there is some natural $b$ so that $y^2 = bx$. Using this, we find that

\begin{align*} x^2 + bx + 1 = xyz. \end{align*} The right-hand side $xyz$ is divisible by $x$, but also $xyz - 1$ is divisible by $x$, since this is equal to $x^2 + bx$. Thus we must have $x = 1$. Now the original equation becomes

\begin{align*} y^2 + 2 &= yz\\ &\Updownarrow \\ y^2 - yz + 2 &= 0. \end{align*} This quadratic equation in $y$ must have a square discriminant since $y$ is assumed to be a natural number. Hence \begin{align*} D &= (-z)^2 - 4\cdot 1 \cdot 2 \\ &= z^2 - 8 \end{align*} is a square. This is only achieved for $z = 3$:

\begin{align*} z^2 - 8 &= r^2 \\ &\Updownarrow \\ (z-r)(z+r) &= 8 \\ &\Downarrow \\ z-r &\in \lbrace 1, 2\rbrace, \\ z+r &\in \lbrace 4, 8 \rbrace. \end{align*} The only possibility here is that $z = 3$ and $r = 1$.

Thus, in any case, $z = 3$.

Answer 2

We can also use Vietas formulas directly to do this problem. Assuming a minimal solution then creating a smaller one if y>x thus y=x is the only possibility.