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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!

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To me it seems clear that there are no solutions, because the difference between those two formulas is just going to get bigger. I am thinking about how to prove this. – Nick Thomas Jun 17 '13 at 4:31
Are you interested just in positive integer solutions, or in arbitrary integer solutions? To me the latter question seems substantially harder. – Nick Thomas Jun 17 '13 at 5:00
my interesting is positive integer. But i think there may be negative solution when n is odd in case there is no positive solution. – chenbai Jun 17 '13 at 5:24
@chenbai : If you want to check some cases higher than $n=41$, I think you can just check until (right side) - (sum on left) becomes positive. After that there won't be solutions. – coffeemath Jun 17 '13 at 12:27
up vote 1 down vote accepted

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:

There may be other links there to further get ideas...

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thanks a lot. it is lucky to know it to avoid many calculations. – chenbai Jun 17 '13 at 23:30

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.)

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but there is always two $x$s, the integers can let RHS or LHS >0 and <0, when n is bigger. – chenbai Jun 17 '13 at 5:30
That's right; but if $x$ and $n$ are big enough, the term $x^{x+n-1}$ overtakes $x^{n+1}$ and $(x+n)^n$. In a polynomial $p(x)$, when $x$ gets big, the term with the largest exponent ends up being the only one that "matters." – Nick Thomas Jun 17 '13 at 5:33
For any specific $n$, the function on the left side of the inequality is negative at $x=1$. Then as the real variable $x$ increases, your argument shows the left side eventually becomes positive. So for any fixed $n$ there is a value (or values) of $x>1$ for which the left side is $0$. The point is to somehow show such $x$ cannot be integral. So I don't think an argument based on an inequality can work. – coffeemath Jun 17 '13 at 5:54
If by some careful approximations one could trap the real $x$ where the sign change occurs, and trap it between integers, it would be done. Experimentally there are very large jumps as one goes (using integer $x$) from negative to positive. – coffeemath Jun 17 '13 at 6:04
Added to last comment-- it's only necessary to trap it relatively near the middle of a gap between adjacent $n$th powers. – coffeemath Jun 17 '13 at 7:02

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