Are there more integral solutions for $3^x+4^y=5^z$, than $x=y=z=2$ ?

If not, how do I show that? I could show that for $3^x+4^x=5^x$, but I'm stuck at the general case? Any ideas, maybe graphs, logarithms or infinite descent?

  • $\begingroup$ $x=0, y=z=1$ is also a solution. Perhaps you mean positive integer solutions? $\endgroup$ Apr 29, 2021 at 3:51

3 Answers 3


I will prove that the only positive integral solution to $3^x+4^y=5^z$ is $x=y=z=2$.

Proof. Looking at the equation mod $4$, we see $3^x\equiv 1\pmod{4}$, or equivalently, $(-1)^x\equiv 1\pmod{4}$. This implies $x=2x_1$ for some integer $x_1$. Also, looking at the equation mod 3, we see $5^z\equiv 1 \pmod{3}$, or equivalently $(-1)^{z}\equiv 1\pmod{3}$. This implies $z=2z_1$ for some integer $z_1$. Thus, $$ 2^{2y}=4^{y}=(5^{z_1})^2-(3^{x_1})^2=(5^{z_1}+3^{x_1})(5^{z_1}-3^{x_1}) $$ Hence, $5^{z_1}+3^{x_1}=2^{s}$ and $5^{z_1}-3^{x_1}=2^{t}$, with $s>t$ and $s+t=2y$. Solving for $5^{z_1}$ and $3^{x_1}$, we get $$ 5^{z_1}=2^{t-1}(2^{s-t}+1) \ \ \textrm{ and } \ \ 3^{x_1}=2^{t-1}(2^{s-t}-1) $$ Since the left side of both equalities is odd, $t$ must be equal to $1$. Let $u=s-t$. Then, the equation $3^{x_1}=2^{t-1}(2^{s-t}-1)$ becomes $3^{x_1}=2^{u}-1$. Looking at this equation mod $3$, we get $0\equiv (-1)^{u}-1\pmod{3}$, and so $u$ is even, say $u=2u_1$ for some positive integer $u_1$. Thus, $$ 3^{x_1}=(2^{u_1})^{2}-1=(2^{u_1}+1)(2^{u_1}-1) $$ Hence, $2^{u_1}+1=3^{\alpha}$ and $2^{u_1}-1=3^{\beta}$ for some $\alpha>\beta$. But this gives, $3^{\alpha}-3^{\beta}=2$, and hence $\alpha=1$ and $\beta=0$. Consequently, $u_1=1$, and so $u=2$. This gives us the unique solution $x=y=z=2$.

  • $\begingroup$ nice solution Prism...... $\endgroup$
    – juantheron
    Nov 20, 2013 at 2:23
  • $\begingroup$ @juantheron: Thanks :) I am glad you liked it! $\endgroup$
    – Prism
    Nov 20, 2013 at 3:30
  • $\begingroup$ thanks! and I think it's just not fair to ask this question in an exam meant for a 12th grade student! $\endgroup$
    – Shubham
    Nov 26, 2013 at 5:56
  • $\begingroup$ I also like this solution. Thanks for taking the time to write it up. $\endgroup$ Apr 29, 2021 at 4:23
  • 1
    $\begingroup$ @Daniel That is a good point! Thanks a lot for the comment. I guess I was implicitly assuming that $x, y, z$ must be positive to have any hope getting a solution. But now, I see that probably more argument is required. EDIT: Oh I see, you already commented on the main post that there are some other solutions when $x, y, z$ are allowed to take $0$ as a value. I've added the assumption on positivity now. $\endgroup$
    – Prism
    May 5, 2021 at 0:31

If a stronger version of $abc$ conjecture is true, then the answer to your question is "no" when $x,y,z>0$.

Statement: If $a+b=c,(a,b)=(a,c)=(b,c)=1,a,b,c>0$ then $$c\leq (rad(abc))^2$$

$rad(n)$ is the product of the distinct prime factors of $n$.

Under the strong version of $abc$ conjecture, $5^z<(3\times 2\times 5)^2$, which means $z\leq 4$, and it is easy to check the other cases.

  • $\begingroup$ Nice perspective! See my answer for elementary solution :) $\endgroup$
    – Prism
    Nov 18, 2013 at 10:22

This plot might help. (A plot of the log base 5.)


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