# About the infinite solutions of a Diophantine equation

Consider the following problem: $$\sum_{k=1}^N k^2=q^2$$ where q is an integer number. This can be written as: $$\frac{1}{3}N^3+\frac{1}{2}N^2+\frac{1}{6}N=q^2$$ In the same way we can write: $$\sum_{k=1}^N k^3=q^3$$ that means: $$\frac{1}{4}N^4+\frac{1}{2}N^3+\frac{1}{4}N^2=q^3$$ In general, we can consider the following equation: $$\sum_{k=1}^N k^r=q^r$$ with $r$ integer number. The question is: does this equation has infinite solutions for every $r$ integer? Thanks.

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$$\frac{1}{4}N^4+\frac{1}{2}N^3+\frac{1}{4}N^2=\left(\frac{N(N+1)}2\right)^2$$ –  lab bhattacharjee May 15 '13 at 11:05
I see. How this could give an answer to the question? –  Riccardo.Alestra May 15 '13 at 11:10
this was just an observation, never an answer –  lab bhattacharjee May 15 '13 at 11:12

Generally, Schaffer proved in 1956 that the equation $$\sum_{k=1}^K k^r = q^n$$ has at most finitely many solutions unless $$(r,n) \in \{ (1,2), (3,2), (3,4), (5,2) \}.$$ The proof uses Siegel's theorem (though, as Erick notes, Faltings would lead to the same conclusion).
Not always,when $r=2$, the equation $x(x+1)(2x+1)=6y^2$ has only these solutions:$(x,y)=(0,0)(-1,0)(1,±1)(24,±70).$
@Riccardo.Alestra By Faltings's Theorem, it is atypical for curves of degree $>2$ to admit infinitely many solutions, so one would guess this is the exact opposite of the truth unless you know of some special structure here. –  Erick Wong May 15 '13 at 14:49