# Infinitely many solutions to $a^x+b^y=c^z$ for given $x, y, z \in \mathbb{N}$?

Problem I have recently seen a few questions such as this(1) and this(2) and this has led me to wonder what conditions are sufficient on $x, y, z \in \mathbb{N}$ so that when given $x, y, z$, $a^x+b^y=c^z$ has infinitely many solutions. This seems like a well known problem, but I have not been able to come across it before so perhaps someone could enlighten me.

Progress: I have already shown that it is sufficient (but by no means necessary) that $x, y, z$ are pairwise coprime.

Proof: Though many proofs can be found, I will write one (rather nice) proof here.

We first notice that $$2^p3^q+2^{p+1}3^q=2^p3^q(1+2)\\ =2^p3^{q+1} \ \forall p, q \in \mathbb{N}$$

It would be great if $2^p3^q, 2^{p+1}3^{q} \ \mbox{and} \ 2^p3^{q+1}$ were powers of $x, y, z$ respectively. However, we can find sufficient conditions on $p, q$ so that such is the case. The following are the sufficient conditions:

• $p,q \equiv 0 \mod{x}$

• $p+1, q \equiv 0 \mod{y}$

• $p, q+1\equiv 0\mod{z}$

Then $2^p3^q$ has an exponent divisible by $x$, $2^{p+1}3^q$ has an exponent divisible by $y$, and $2^p3^{q+1}$ has an exponent divisible by $z.$

In other words, if we can find infinitely many $p$ and $q$ so that the three above conditions are satisfied, we have found infinitely many solutions to $a^x+b^y=c^z$ given $x, y, z \in \mathbb{N}$ pairwise coprime. But the conditions are equivalent to the following:

$$p\equiv \begin{cases}0 \ \ \ \mod{x} \\ 0 \ \ \ \mod{z} \\-1 \mod{y} \end{cases}$$ and $$q\equiv \begin{cases}0 \ \ \ \mod{x} \\ 0 \ \ \ \mod{y} \\-1 \mod{z} \end{cases}$$

But since $x, y, z$ are pairwise coprime we can use the Chinese Remainder Theorem, giving us unique solutions for $p$ and $q$ modulo $xyz$. In particular, we have found infinitely many values of $p$ and $q$, which is what we wanted.

By no means are these conditions necessary, consider the simple case $a^2+b^2=c^2$ which is known to have also infinitely many solutions, but in this case $x, y, z$ are not pairwise coprime. So for what other $\langle x, y, z \rangle$ gives rise to infinitely many solutions to $a^x+b^y=c^z$?

• It only suffices that $\gcd(z,xy)=1$; but it is not necessary. – Davood KHAJEHPOUR Aug 23 '17 at 10:20

If $\gcd(z,xy)=1$, then by chinese reminder theorem; there is infinitly many $k$ such that:

$$k \overset{xy}{\equiv} 0 \ \ \ \ \text{and} \ \ \ \ k \overset{z}{\equiv} -1;$$

now let $$a:=2^{ \tiny{ \dfrac{k} {x} } }, \ \ \ \ \ \ b:=2^{ \tiny{ \dfrac{k} {y} } }, \ \ \ \ \ \ c:=2^{ \tiny{ \dfrac{k+1} {z} } };$$

and notice that: $\ 2^k+2^k=2^{k+1}$.

Let's consider the equation:

$$x^p+y^q=z^r;$$

with $2 \leq p, q, r$.

• If $\ \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} \leq 1$ ;
then there is no polynomial parametric solution for coprime $x,y,z$.

• If $\ 1 < \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r}$ ;
then there is polynomial parametric solution for coprime $x,y,z$.
This case contains the following cases:
$$\ \ \ \ \ \ \ \ \ \ \ \{ p,q,r \} = \{ 2,3,6 \}, \ \ \ \{ p,q,r \} = \{ 2,3,5 \}, \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \{ p,q,r \} = \{ 2,3,4 \}, \ \ \ \{ p,q,r \} = \{ 2,3,3 \}; \ \ \ \\ \{ p,q,r \} = \{ 2,2,n \}, \ \ \ \ \ \ \ \ \ \text{for every} \ n .$$

• Ah thanks for your answer and for improving the set of solutions :-) – Jihoon Kang Aug 23 '17 at 12:56
• But if $\{p,q,r\}=\{2,3,6\}$ then $\tfrac1p+\tfrac1q+\tfrac13=1$. Should this combination not be included, or are the inequalities incorrect? – Servaes Jun 15 '19 at 16:38