<s>[INST] Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$ 
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$

It's easy to see that $x=0$ and $x=1$ are solutions but are these the only one? How do I demonstrate that?
I've tried to write them either:
$$5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}$$
or
$$5^x+7^x+11^x=(5+1)^x+(7+1)^x+(11-2)^x$$
and tried to think of some AM-GM mean inequality or to divide everything by $11^x$, but those don't seem like the way to go. Any hints?
 [/INST] You write the equation in the form
$$
[(9+2)^k-9^k]-[(7+1)^k-7^k]-[(5+1)^k-5^k]=0.
$$
If $k\ge 0$, then you can use that 
$$
(9+2)^k-9^k=\int_9^{9+2}kt^{k-1}dt\ge 2k9^{k-1},\\
-[(7+1)^k-7^k]=-\int_7^{7+1}kt^{k-1}dt\ge -k8^{k-1},\\
-[(5+1)^k-5^k]\ge -k6^{k-1}.
$$
Your equation becomes
$$
0\ge k\cdot\left(2\cdot 9^{k-1}-8^{k-1}-6^{k-1}\right).
$$
If we discard the $k=0$ solution, it suffices to examine the inequality
\begin{align}
6^{k-1}+8^{k-1}&\ge 2\cdot 9^{k-1},\qquad k=1,2,3,4,...
\end{align}
This is equivalent to the inequality
$$
6^\ell+8^\ell\ge 2\cdot 9^\ell,\qquad \ell=0,1,2,3,...,
$$
if $k=\ell+1$.  Observe that
$$
9^\ell +9^\ell\ge 6^\ell+8^\ell\ge 2\cdot 9^\ell
$$
since $\ell\ge 0$.  This means equality holds, and the only way that can be true is if $\ell=0$.  (to see this step, rewrite as $0<9^\ell-6^\ell=-(9^\ell-8^\ell)<0$.)
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