Solving the equation $yz=y^z-2$

Question

Question: Let $x,y$ and $z$ be positive integers. How can we determine the positive integer solutions to the equation $$yz=y^z-2$$

The motivation here is straightforward. Surely $6=2\times 3$ and also $8=2^3$ and $9=3^2$. I wanted to "generalize" those observations by writing

\begin{align} x&=yz\\ x+2&=y^z\\ x+3&=z^y \end{align}

Trivially we can get to $y^z-2=x$ and after substitution we have the equation $$yz=y^z-2$$

WolframAlpha quickly spits out $y=2$ and $z=3$ as the only positive integer solutions. It also gives a "solution for the variable z" as

$$z=-{yW\left(-y^{-1-{2\above 1.5pt y}} \right)-2\log{y}\above 1.5pt y\log{y}} $$

where $W(a)$ is the Lambert W function. Following the Wiki link leads me to think a possible start is to consider $$-t=x+{2\above 1.5pt y} $$ and we can now transform our equation into $$ty^t=-{1\above 1.5pty}y^{-{2\above 1.5pt y }}$$ which apparently yields the solution for $z$ written above. But I am still not clear how WolframAlpha explicitly determined that $z=3$.

Answer 1

Lemma(I): Let $3 \leq y$, then we have:
$$2y < y^2-2.$$

Proof: $3 \leq y$, so we must have:

$$2 \leq (y-1) \Longrightarrow 3 < 4 \leq (y-1)^2 \Longrightarrow 3 < y^2-2y+1 \Longrightarrow 2y < y^2-2 . $$

---

Lemma(II): Let $3 \leq y$ be any fixed arbitrary integer and let $2 \leq z$, then we have: $$ \color{Red}{yz < y^z-2} \ \ \ \ \ \ \ \ \ \ \text{(II)} . $$

Proof: Let $y$ be an arbitrary (but fixed) integer greater or equal than $3$. We will prove the lemma by induction on $z$.

$z=2$; which is done by Lemma(I).

Now suppose that the Lemma(II) is true for $z=k$; then we will show it for $z=k+1$. Only notice that:

$$ \color{Blue}{y} < zy+2 < y^z < (y-1)y^z = \color{Blue}{y^{z+1}-y^z} \ ; $$

adding this last inequality by inequlaity (II) we get:

$$ y(z+1)= \color{red}{yz}+ \color{Blue}y < \color{Red}{y^z-2} + \color{Blue}{y^{z+1}-y^z} = y^{z+1}-2 .$$

---

Lemma(III): Let $y=2$ and let $4 \leq z$, then we have: $$ \color{Red}{2z < 2^z-2} \ \ \ \ \ \ \ \ \ \ \text{(III)} . $$

Proof:
We will prove the lemma by induction on $z$.

$z=4$; is trivial.

Now suppose that the Lemma(III) is true for $z=k$; then we will show it for $z=k+1$. Only notice that:

$$ \color{Blue}{2} < 2z+2 < 2^z = \color{Blue}{2^{z+1}-2^z} \ ; $$

adding this last inequality by inequlaity (III) we get:

$$ 2(z+1)= \color{red}{2z}+ \color{Blue}2 < \color{Red}{2^z-2} + \color{Blue}{2^{z+1}-2^z} = 2^{z+1}-2 .$$

---

---

---

---

First Case: Let $3 \leq y$:

• $2 \leq z$; which is impossible by Lemma(II).
• If $z=1$; then we have $y.1=y^1-2$, which is obviously impossible again!

Second Case: Let $y=2$:

• $4 \leq z$; which is impossible by Lemma(III).

• $z=3$; which gives the solution $\color{Green}{(y,z)=(2,3)}.$

• $z=2$; which does not give any solution!

• $z=1$; which does not give any solution again!

Third case: Let $y=1$:

• in this case we have: $1.z=1^z-2$; which does not have a solution in $\mathbb{N}$.