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Finding all real number $n$ in

$$5^n+7^n+11^n=6^n+8^n+9^n$$

Try: From given equation

$n=0,1$ are the solution

But i did not understand any other solution exists or not

Although i have tried like this way

$$\bigg(\frac{5}{9}\bigg)^n+\bigg(\frac{7}{9}\bigg)^n+\bigg(\frac{11}{9}\bigg)^n = \bigg(\frac{6}{9}\bigg)^n+\bigg(\frac{8}{9}\bigg)^n+1$$

Right side is strictly increasing function. but i have a confusion whether left side is strictly increasing or not

could some help me how to solve it, thanks

Community
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DXT
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    It is not correct that the right hand side of your equation is strictly increasing in $n$. Hint: can you say something about the maximum size of the right hand side? – Gnuk Nov 21 '18 at 08:40
  • See also here: https://math.stackexchange.com/questions/2840394 – Michael Rozenberg Nov 21 '18 at 12:09

1 Answers1

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Consider the function $f(x)=x^n$ for positive $x$. Its second derivative is $n(n-1)x^{n-2}$ and therefore for $n > 1$ or $n<0$ $f$ is strictly convex while for $0 < n < 1$ f is strictly concave.

Our equation is equivalent to $f(5) + f(7) + f(11) = f(6) + f(8) + f(9)$

In the convex case we have:

$f(6) = f(\frac{5+7}{2}) < \frac{f(5)+f(7)}{2}$,

$f(9) = f(\frac{7+11}{2}) < \frac{f(7)+f(11)}{2}$,

and that

$f(8) = f(\frac{11+5}{2}) < \frac{f(11)+f(5)}{2}$

So $ f(5) + f(7) + f(11) > f(6) + f(8) + f(9)$

In the concave case the same strict inequalities hold, but reversed. Thus only n=0 or n=1 could be solutions and they both work.

Sorin Tirc
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