For 17 integers: $a_{i} (i=1,2,\dotsm,17)$, $a_{1}^{a_{2}}=a_{2}^{a_{3}}=a_{3}^{a_{4}}=\dotsm=a_{16}^{\quad a_{17}}=a_{17}^{a_{1}}$.

Question

Full question: For 17 integers: $a_{i} (i=1,2,\dotsm,17)$, $a_{1}^{a_{2}}=a_{2}^{a_{3}}=a_{3}^{a_{4}}=\dotsm=a_{16}^{\quad a_{17}}=a_{17}^{a_{1}}$. Prove that $a_{1}=a_{2}=a_{3}=\dotsm=a_{16}=a_{17}$.

I know this does not hold true for i=1,2 (where there are only two terms), so I'm thinking if there's some relationship which I missed. Any help would be appreciated.

Answer 1

Begin by looking at the simple case $a_1^{a_2}=a_2^{a_1}$. You know the only non-trivial solution (i.e. $a_1\ne a_2$) in the integers is $(a_1,a_2)=(2,4)$, up to order. Let's begin by saying $2^4=4^2$. Now add one more term, $a_3$, which now must be the exponent of $a_2$, which is $2$. You get $2^4=4^2=2^{a_1}$. But $a_1=2$, so this statement cannot be true. If you add a fourth term, you get $2^4=4^2=2^4=4^{a_1}$. In this case, since $a_1=2$, the statement is true again.

You can see that the chain of proposed equalities can be true only if it contains an even number of integers. But $17$ is odd, so it cannot be true for the non-trivial solutions to the base case $2,4$. Therefore, for chains of odd length, it is true only in the trivial case that $a_i=a_j$ in every instance.

Answer 2

Hint If $x^y=y^x$ then $$\frac{\ln(x)}{x} = \frac{\ln(y)}{y}$$

Now, $f(x)=\frac{\ln(x)}{x}$ is increasing on $(0,e)$ and decreasing on $(e , \infty)$.

Deduce from here that $x^y=y^x$ implies that exactly one of the following happens

• $x=y$
• $x<e<y$
• $y <e<x$.

Now, assume by contradiction that $a_1 \neq a_2$. Show first that $a_{j} \neq a_{j+1}$ for all $1 \leq j \leq 16$.

Next, deduce that if $a_j <e$ then $a_{j+1}>e$ and if $a_{j} >e$ then $a_{j+1}<e$. Compare $a_1$ with $a_{17}$.