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Consider a sequence $(a_n)$ of positive numbers such that $$(a_1+\cdots+a_n)^2=a_1^3+\cdots +a_n^3,\quad n\ge 1.$$ Prove that $a_n=n$ for all $n\ge 1$.

Ana María CG
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    Did you try proving that using induction? – GBQT Nov 25 '15 at 12:32
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    Welcome to math stack exchange! The implication is only true for increasing sequences. Otherwise, the sequence {5,3,4,1,2} would be a counterexample. – Peter Nov 25 '15 at 12:55
  • @Peter I don't think (5,3,4,1,2,...) is a counterexample; it does not satisfy the condition on the sequence for n = 2, because $5^3 + 3^3 \neq (5 + 3)^2$. – user19610 Sep 15 '22 at 20:21

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The question should specify that the sequence is increasing since $\{1,2,4,3,5\}$ is a counterexample.

Let's prove this by induction. The base case of $a_1 = 1$ holds since $a_1^2 = a_1^3$. Now assume that $(a_1,\ldots,a_k) = (1,\ldots,k)$ for some $k$. Then we have that $(1+2+\cdots+k+a_{k+1})^2 = \Bigg(\dfrac{(1+k)k}{2}+a_{k+1}\Bigg)^2$. Also, we have that $1^3+2^3+\cdots+a_{k+1}^3 = \dfrac{(1+k)^2k^2}{4}+a_{k+1}^3$. Then, setting them equal we obtain $\Bigg(\dfrac{(1+k)k}{2}+a_{k+1}\Bigg)^2=\dfrac{(1+k)^2k^2}{4}+a_{k+1}^3 \longrightarrow a_{k+1}(k^2+k)+a_{k+1}^2-a_{k+1}^3 = 0 \longrightarrow a_{k+1}(1+k-a_{k+1})(k+a_{k+1}) = 0$.

Since the sequence is positive, out of the solutions $a_{k+1} = 0, -k,$ and $k+1$, only $k+1$ works thus finishing the induction.

user19405892
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    Why do you need the increasing hypothesis, since you know that the sequence is positive, leaving only $k+1$ as the solution? – GBQT Nov 25 '15 at 13:33
  • I was just using the fact that since $a_1 = 1$ and the sequence is increasing, $k+1$ must be the solution. But you could also obtain the same result by saying the sequence is positive. – user19405892 Nov 25 '15 at 13:41
  • @user19405892, I don't think $(1,2,4,3,5,...)$ is a counterexample; it does not satisfy the condition on the sequence for $n=3$, as $1^3 + 2^3 + 4^3 \neq (1 + 2 + 4)^2$. – user19610 Sep 15 '22 at 20:24