It is considered arithmetic progression $x_1,x_2, \cdots, x_n,\cdots, x_1 \neq0$ . Show that if sums $$S_n = x_1^3+x_2^3+ \cdots +x_n^3$$ is squares perfect for any natural $n \in N$, then there are $k\in N^*$ so $x_n=nk^2$, for any $n \in N$. All my attempts were fruitless.
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1What is $N^*$ ? – 8hantanu Jan 21 '17 at 19:40
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@WiCK3DPOiSON :$N^*$ is the set of integers strictly positive. – medicu Jan 22 '17 at 07:02
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Do we assume that all $x_i \in \mathbb N$? – Mariuslp Jan 22 '17 at 15:59
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1I think so, because otherwise we can take $x_1=0$ and $x_{n+1}-x_n=k^2$. For this progression it is easy to see that all $S_n$ are square numbers, but $x_n=(n-1)k^2$ and the problem would be false, say for choosing $k=3$. – Shroud Jan 22 '17 at 19:41
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I'm not sure if this helps, but note that $$\sum_{m=1}^{n}{m^3}={(\frac{n(n+1)}2)^2}$$ Setting $x_i=nk^2$ we get $$S_n=\sum_{m=1}^{n}{(mk^2)^3}= k^6\sum_{m=1}^{n}{m^3}={(\frac{k^3n(n+1)}2)^2}$$ This is only the converse, however. I have no idea how to proceed for the forward direction.
woogie
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