5
$\begingroup$

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?

$\endgroup$
15
$\begingroup$

I will prove that the only 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 $x=y=z=2$.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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.