8
$\begingroup$

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?

$\endgroup$
2
20
$\begingroup$

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 $x=y=z=2$.

$\endgroup$
6
  • $\begingroup$ nice solution Prism...... $\endgroup$
    – juantheron
    Nov 20 '13 at 2:23
  • $\begingroup$ @juantheron: Thanks :) I am glad you liked it! $\endgroup$
    – Prism
    Nov 20 '13 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 '13 at 5:56
  • $\begingroup$ I also like this solution. Thanks for taking the time to write it up. $\endgroup$ Apr 29 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 at 0:31
5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Nice perspective! See my answer for elementary solution :) $\endgroup$
    – Prism
    Nov 18 '13 at 10:22
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.