# Positive integral solutions of $3^x+4^y=5^z$

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?

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

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 $$x=y=z=2$$.

• nice solution Prism...... Nov 20, 2013 at 2:23
• @juantheron: Thanks :) I am glad you liked it! Nov 20, 2013 at 3:30
• thanks! and I think it's just not fair to ask this question in an exam meant for a 12th grade student! Nov 26, 2013 at 5:56
• I also like this solution. Thanks for taking the time to write it up. Apr 29, 2021 at 4:23
• @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. 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.

• Nice perspective! See my answer for elementary solution :) Nov 18, 2013 at 10:22

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