Does $\sum_{k=1 }^ n (x-1+k)^n=(x+n)^n$ have integer solution when $n\ge 4$? from a post in SE, one says $3^2+4^2=5^2,3^3+4^3+5^3=6^3$,that is interesting for me. so I begin to explore further, the general equation is  
$\sum_{k=1}^n (x-1+k)^n=(x+n)^n$, 
from $n \ge 4$ to $n=41$, there is no integer solution for $x$. 
for $n>41$,I can't get result as Walframalpha doesn't work.
I doubt if there is any integer solution for $n \ge 4$, when $n$ is bigger,the $x$ is close to $\dfrac{n}{2}$.
Can some one have an answer? thanks!
 A: Your problem has been studied, and it is conjectured that only $3,4,5$ for squares and $3,4,5,6$ for cubes are such that all the numbers are consecutive, and the $k$th power of the last is the sum of the $k$th powers of the others ($k>1$).
I've spent a lot of time on this question, and no wonder did not get anything final on a proof, since apparently nobody else has succeeded in proving it.
The website calls it "Cyprian's Last Theorem", arguing that it seems so likely true but nobody has shown it yet, just like for Fermat for so many years.
The page reference I found:
http://www.nugae.com/mathematics/cyprian.htm
There may be other links there to further get ideas...
A: Here is an argument that there are no solutions when $x$ is positive and $n$ and $x$ are sufficiently large. What "sufficiently large" means, I do not know exactly. I am making this answer community wiki; maybe others can improve on it.
$$ (\sum_{k=x}^{x+n-1} k^n) - (x+n)^n \geq \int_x^{x+n-1} t^n dt - (x+n)^n  = \frac{1}{n+1}(x^{x+n-1} - x^{n+1}) - (x+n)^n, $$
which is greater than zero if $x$ and $n$ are sufficiently large. (Note that I wrote your initial sum a little differently than you did.)
