$\ell_p$ is Hilbert space if and only if $p=2$

Let p greater than or equal to 1,show that the space of all p-summable sequences is an inner product space if and only if p=2

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Hint: Parallelogramm law, $e_1$, $e_2$. –  martini Oct 18 '12 at 13:10
Hint: Prove that if $p=2$, the parallelogram law is satisfied. WHen $p\neq 2$ give a counter example to the parallelogram law . –  Tomás Oct 18 '12 at 13:11
This question gives more details on parallelogram law: Norms Induced by Inner Products and the Parallelogram Law –  Martin Sleziak Oct 18 '12 at 13:15
I think the question is whether $\ell_p$ admits an inner product making $\ell_p$ to a Hilbert space. The corresponding norm doesn't have to be $\Vert\cdot\Vert_p$ (an equivalent norm would do). None of the comments above seems to answer this question. –  user8268 Oct 18 '12 at 13:19
related: How do you prove that $\ell_p$ is not isomorphic to $\ell_q$? and the more general answer in If $1\leq p \lt \infty$ then show that $L_p([0,1])$ and $\ell_q$ are not topologically isomorphic. –  user45115 Oct 18 '12 at 14:16

Assuming we are working with the usual norm (as Op said in comments), suppose $\ell_{p}$ is an Hilbert space. So its must satisfy for all $u,v$: $$2||u||_{p}^2 + 2||v||_{p}^2 = ||u + v||_{p}^2 + ||u - v||_{p}^2$$.
As suggested by martini, take $u=e_{1}=(1,0,...,0,...)$ and $v=e_{2}=(0,1,0,...,0,...)$. Hence, by the last equality, we have $$4=2^{\frac{2}{p}}+2^{\frac{2}{p}}$$
Now you can solve the last inequality and verify that $p=2$.
On the other hand, if $p=2$, you can easily check that $\ell_{2}$ is a Hilbert space.