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I'm looking for a hint at this problem. I've thought about induction but it didn't work.

Let $n$ be a natural number greater or equal to $2$. Prove that the equation has an infinite number of solutions in ℕ* $$a_2^{2} + a_3^{3} + ... + a_n^{n} = a_{n+1}^{n+1}$$

For $n=2$, we get $a_3=\sqrt[3]{a_2^2}$

for $n=3$,
$$a_2^{2} + a_3^{3} = a_4^{4}$$ $$2a_2^{2}=a_4^{4}$$ $$a_4 = \sqrt[4]{2a_2^{2}}$$

for $n=4$, we get $a_5 = \sqrt[5] {4a_2^{2}}$

for $n=5$, we get $a_6 = \sqrt[6] {8a_2^{2}}$

So, following this rule, $a_n = \sqrt[n] {xa_2^2}$, where $x$ is a power of $2$? But does it help? Anyway, I think it needs a totally different approach.

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    Related : [Is there a systematic way to solve in $\bf Z$: $x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$ for all $n$?](http://math.stackexchange.com/questions/1619089/is-there-a-systematic-way-to-solve-in-bf-z-x-12x-23-x-nn1-zn) – 2017-02-17

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