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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?

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  • $\begingroup$ From given equation no other solution other then $x=0,1$ but unable to prove it, $\endgroup$ – DXT Jul 4 '18 at 7:39
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Consider the function for $x>1,n>1$

$$f(n)=n^x-(n-1)^x$$ $$f'(n)=xn^{x-1}-x(n-1)^{x-1}=x(n-1)^{x-1}[(1+\frac{1}{n-1})^{x-1}-1]>0$$

So $f(n)$ is increasing for $x>1,n>1$.

Now rewrite the equation $$5^x+7^x+11^x=6^x+8^x+9^x$$ $$\color{red}{(11^x-10^x)}+\color{blue}{(10^x-9^x)}=\color{red}{(8^x-7^x)}+\color{blue}{(6^x-5^x)}$$ Comparing the red and blue parts, LHS is larger than RHS due to increasing $f(n)$.

So the equation holds only if $0\le x\le 1$ for natural number solutions, i.e. $x=0$ or $x=1$.

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  • $\begingroup$ diophantine solutions? $\endgroup$ – C. Cristi Jul 4 '18 at 10:02
  • $\begingroup$ Integer solutions. $\endgroup$ – Mythomorphic Jul 4 '18 at 10:05
  • $\begingroup$ Can you be more detalied why is LHS is larger than RHS then the equation only hold for $0\leq x \leq1$? $\endgroup$ – C. Cristi Jul 4 '18 at 10:09
  • $\begingroup$ Since $f(n)$ is increasing for $x>1, n>1$, it follows that $f(11)>f(8)\iff11^x-10^x>8^x-7^x$ for all $x>1$. LHS is larger than RHS in this case. If we want to see LHS=RHS, the only possible case remaining is $x\le 1$. Since you are seeking for natural solution of $x$, $x$ can either be $0$ or $1$. Same argument for the blue part. $\endgroup$ – Mythomorphic Jul 4 '18 at 10:16
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Let $f(x)=x^k,$ where $k>1$ or $k<0$.

Thus, $f$ is a convex function and since $(11,7,5)\succ(9,8,6),$ by Karamata we obtain: $$f(11)+f(7)+f(5)>f(9)+f(8)+f(6).$$

Also, for $0<k<1$ we see that $f$ is a concave function.

Thus, by Karamata again $$f(11)+f(7)+f(5)<f(9)+f(8)+f(6).$$ Thus, it remains to check, what happens for $k\in\{0,1\}$.

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Another elemenatary solution, using that $x$ is supposed to be a natural number. It even uses a technique the OP considered:

The equation is equivalent to

$$\left(\frac{5}{11}\right)^x + \left(\frac{7}{11}\right)^x +\left(\frac{11}{11}\right)^x = \left(\frac{6}{11}\right)^x +\left(\frac{8}{11}\right)^x +\left(\frac{9}{11}\right)^x$$

All terms are positive and the left hand side contains a summand 1 in the form of $\left(\frac{11}{11}\right)^x$.

From the terms on the right hand side, $\left(\frac{9}{11}\right)^x$ is the biggest, but it will of course still decrease for increasing $x$. Using a calculator will show you that $\left(\frac{9}{11}\right)^6 < \frac13$.

That means for $x \ge 6$, the right hand side consists of the sum of 3 values, the highest of which is less than $\frac13$. That means the right hand side is less than 1, while the left hand side is bigger than 1, leading to a contradiction.

Now 'only' the cases $x=2,3,4,5$ need to be checked by hand, and they don't lead to the equation being fullfilled.

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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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    $\begingroup$ I don't quite understand how you get that the only solutions must be $0,1$. I understand FTC and why you wrote them like that but can you add more detalies? $\endgroup$ – C. Cristi Jul 4 '18 at 10:14
  • $\begingroup$ I added some more details, hope it helps. $\endgroup$ – user254433 Jul 4 '18 at 21:02

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