Existence of Polynomial with whom product contains only prime powers Let $f \in \mathbb{Q}$ be a polynomial of degree $n>0$. Let $p_1,p_2,p_3..p_{n+1}$ be distinct prime numbers. Show that there exists a non-zero polynomial $g \in \mathbb{Q}[X]$ such that $fg=\sum_{i=1}^{n+1}c_iX^{p_i} $ with $c_i \in \mathbb{Q}$.
If I want to find a polynomial then it must be of degree $p_{n+1}-n$.
So I assumed a polynomial $g=\sum_{i=1}^{p_{n+1}-n}b_iX^{i}$.
Multiplying $f$ & $g$ and equating the non-prime power coefficents to $0$ where our variables are $b_i's$ (they are $p_{n+1}-n+1$ in number) and we have $p_{n+1}-n+1$ equations.
One solution is all $b_i=0$ and in this case it is zero polynomial. 
How do I proceed further? Is there another way to prove it?

Consider $\Phi: \mathbb{Q}_{p_{n+1}-n}[X] \to \mathbb{Q}_{p_{n+1}}[X]$ defined by letting $\Phi(g) = fg$. 
Now let $V = Vect (X^{p_1},...,X^{p_{n+1}})$ (I'm assuming $p_1<...<p_{n+1}$ hence $V$ is a subvector space of $\mathbb{Q}_{p_{n+1}}[X]$)
As the $p_i$ are distinct, $\mathrm{dim}V = n+1$.
Moreover, $f$ is nonzero so $\Phi$ is injective, so its image has dimension $p_{n+1}-n+1$. 
Now assume the conclusion is not met, this means $\mathrm{Im}\Phi \cap V = \{0\}$. This in turn implies that $V$ is included in a subspace $S$ such that $\mathrm{Im}\Phi \oplus S = \mathbb{Q}_{p_{n+1}}[X]$. This implies $n+1= \mathrm{dim}V\leq \mathrm{dim}\mathbb{Q}_{p_{n+1}}[X] - \mathrm{dim}\mathrm{Im}\Phi = p_{n+1}+1 - (p_{n+1} -n +1) = n$, a contradiction. 
Hence $\mathrm{Im}\Phi \cap V \neq \{0\}$, so there is $g$ such that $fg= \Phi(g) \neq 0$ (in particular $g\neq 0$) and $fg \in V$, so $fg = \displaystyle\sum_{i=1}^{n+1}c_i X^{p_i}$ for some $c_i \in \mathbb{Q}$
Note that this has nothing to do with prime numbers. 
