I'm coaching my student for the next IMO (Spain, July 2008), and we were discussing the following problem, probably some of you already know it, but I think it was interesting.

Find all positive integers n such that:

3^n + 4^n + 5^n + ....... + (n+2)^n = (n+3)^n

Is it true that

lim_{n \to\infty} (RHS/LHS) = e-1 > 1 ?

That would make the number of solutions finite. Probably wrong approach. But it suggests show RHS > LHS for all n > some N.

I got two solutions, then it seems inequality holds.

There are only 2 solutions when n is prime (consider mod n).

There is only 1 solution when n+1 is prime (consider mod n+1).

But that won't complete it.

It looks like an extension of the power means.

I should come up with a solution soon.

There are several ways to see that as n -> infinity, RHS > LHS
but the key idea is to get the induction to work.

Let S(n) be the LHS.

We would like to find n big enough so that:
If S(n)<(n+3)^n, then S(n+1)<(n+4)^(n+1).

S(n+1)= 3^(n+1) + 4^(n+1) +...+ (n+2)^(n+1) +(n+3)^(n+1)
< (n+2)(3^n) + (n+2)(4^n) +...+ (n+2)((n+2)^n) + (n+3)((n+3)^n))
=(n+2)S(n) + (n+3)((n+3)^n))
<(n+2)((n+3)^n) + (n+3)((n+3)^n))
=(2n+5)((n+3)^n))

For this last expression to be <=(n+4)^(n+1), this is equivalent
to:

(2n+5)/(n+4) <= ((n+4)/(n+3))^n = (1 + (1/(n+3)))^n or
1+(n+1)/(n+4) <= (1 + (1/(n+3)))^n

The RHS of the above approaches e>2 as n->infinity, so for
sufficiently large n, a strict inequality holds. The RHS can be
expanded by the binomial theorem to
1 + n/(n+3) + n(n+1)/[2(n+3)^2] + ...
so it is sufficient if n>=3 (details left out).

[ Seems that there should be a more elegant approach than
the above! ]

As a check, S(2) = 25 = 5^2 so equality holds for n=2 and
S(3) = 216 = 6^3 so equality also holds for n=3. For n=4,
S(4) = 2258 < 7^4 = 2401. From the above, since the LHS
is smaller, S(n)<=(n+3)^n for n>=4.