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!

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.)
