Determine all positive integers n such that, for any natural number m ≥ 1, the following equality holds:

[chibyston]

Determine todos os inteiros positivos n tais que, para qualquer número natural m ≥ 1, a seguinte igualdade vale:

[math]\left(1+2^n+3^n+...+m^n\right) = \left(1+2+3+...+m\right)^{n-1}[/math]

 

[BigBeachBanana]

Google translation:

Determine all positive integers n such that, for any natural number m ≥ 1, the following equality holds

 

[Dr.Peterson]

Determine todos os inteiros positivos n tais que, para qualquer número natural m ≥ 1, a seguinte igualdade vale:View attachment 35646

Where does this come from? What have you tried?

To make a guess, you might try finding what values of n will work just for m=2, say. Then try proving that your answer works for any m.

 

[BigBeachBanana]

I tested a few cases from n=0 to 5, it appears that the equality holds for n=3. I believe this is the only solution given the order of growth of the LHS vs. the RHS.

The LHS sum is particularly challenging to resolve. One idea that comes to mind is to create some sort of upper/lower bounds similar to the squeeze theorem.

 

[gonshonmargo]

thank for answers

 

[Dr.Peterson]

I just recognized the formula for a sum of a sequence of cubes, so I knew that would be an answer; then I used the thinking I suggested above to convince myself it is the only answer. I have seen a number of ways to derive the formula, none of which are immediately obvious. But given that we just need to prove a claimed equation, induction might be the best choice. Or not.