Li4 wrote: Find all tuples of positive integers $(a,b,c)$ such that $$a^b+b^c+c^a=a^c+b^a+c^b$$ Any solution?

If I'm not mistaken, a bounding solution works

Bump for solution.

Please correct me if the following solution it is wrong: 1) Consider the functions $f(x)=x^b+c^x+b^c$, and $g(x)=x^c+b^x+c^b$ 2) We can see easy, that both functions there are convex for all positive integer $b,c,x$ 3) So, if the equation $f(x)=g(x)$ have maximum two solutions,and because $x=b$, and $x=c$ there are solutions, these are all the solutions (in this situation $f$ and $g$ there are the same). 4) On the basis of the previouse, $x=a$ can be solution, only if $a=b$ or $a=c$ So all the solutions of the equations there are $(k,k,l)$ and their permutations, and these there are indeed solutions.

^ you forgot the case where $b=c$, in which case $f$ and $g$ are the same. Also, your final conclusion is wrong; you could have solutions like $(k,k,l)$ or $(l,k,l)$.

There exists stronger version of this problem in BMO Shortlist 2014(which is N5)

biomathematics wrote: ^ you forgot the case where $b=c$, in which case $f$ and $g$ are the same. Also, your final conclusion is wrong; you could have solutions like $(k,k,l)$ or $(l,k,l)$. Thank you, my wrong solution can be corrected! So I corrected my demosntration!

how can i add subscript while adding a question?

TuZo wrote: ... 2) We can see easy, that both functions there are convex for all positive integer $b,c,x$ 3) So, if the equation $f(x)=g(x)$ have maximum two solutions,... I don't think so,what if $f(x)=x^4$ and $g(x)=x^2$? (Please tell me if I misunderstand your solution,here I think you deduce #3 from #2 which is not correct.)

Funwith-math wrote: how can i add subscript while adding a question? use underscore. For example, if you write a_i enclosed within dollar signs, you get $a_i$. If the subscript has more terms, use {} brackets; for example, enclosing a_{ijk} will give $a_{ijk}$. More complicated stuff such as $\sum_{i=1}^n \sum_{j=1}^n a_i b_j$ can be written too! For the latex code, you can click on the latex part too, for example $\textbf{CLICK ME HERE!}$. Also, welcome to AoPS Rules are strict here; the question you asked does not belong to the thread, and must be asked where it should be asked.

TLP.39 wrote: TuZo wrote: ... 2) We can see easy, that both functions there are convex for all positive integer $b,c,x$ 3) So, if the equation $f(x)=g(x)$ have maximum two solutions,... I don't think so,what if $f(x)=x^4$ and $g(x)=x^2$? (Please tell me if I misunderstand your solution,here I think you deduce #3 from #2 which is not correct.) Yes, your counterexample is correct,...so how many solution can have the equation $f(x)=g(x)$, have you any suggestion? It is true, that the maximum number of the solutions there are $3$?

quangminhltv99 wrote: Bump for solution. Can you post your solution? Thanks a lot?!

If any two of $a, b, c$ are equal then we have a solution. Now assume without loss of generality that $a < b < c$. For $b\geq 3$ we claim that $a^b+b^c+c^a > a^c+b^a+c^b$. We wish to show that $b^c > c^b+a^c$. First of all, $a\leq b-1$ so it suffices to show $b^c > c^b+(b-1)^c$. Next, we claim that proving the inequality for $c$ implies the inequality for $c+1$. If $b^c > c^b+(b-1)^c$, then \[b^{c+1}\geq b(c^b+(b-1)^c) = bc^b + b(b-1)^c > (c+1)^b +(b-1)^{c+1}.\]The last step relies on \[\left(\frac{c+1}{c}\right)^b\leq\left(\frac{b+2}{b+1}\right)^b < e < b\implies bc^b > (c+1)^b.\]Therefore we may assume $c=b+1$ and we are left to prove the one-variable inequality $b^{b+1} > (b+1)^b+(b-1)^{b+1}$. We have \[\frac{(b+1)^b+(b-1)^{b+1}}{b^{b+1}}=\frac{1}{b}\left(\frac{b+1}{b}\right)^b+\left(\frac{b-1}{b}\right)^{b+1} < \frac{e}{b}+\frac{1}{e} < 1\]for $b\geq 5$. We can check $b=3, 4$ by hand. Given that $b^c > c^b+a^c$, combining with $c^a > b^a$ and $a^b > 0$ yields $a^b+b^c+c^a > a^c+b^a+c^b$ as desired. Now we are left to check $b=2$ (therefore $a=1$), where the equation is $1+2^c+c = 1+2+c^2$. The left hand side grows much faster and it is not hard to argue that $c=3$ is the only solution. In summary, solutions are permutations of $(1, 2, 3)$, or two of $a, b, c$ are equal.