I remember posting it in the "Analysis" section (that was a bad choice ), so try searching for it. Actually, if I'm not mistaken, it was discussed at least twice.

WHY ?????? you said Quote: I remember posting it in the "Analysis" section (that was a bad choice ), I know that it's for the Romanain olympiad

We first look at the case $a\ne 0$. Then all $f_k(1)$ are non-null, and $a=\prod_kf_k(1)$. We divide both sides by $a$, and if we write $g_k$ for $\frac{f_k}{f_k(1)}$, we get $\prod_kg_k(x)=x^n,\ \forall x$. Now fix any $\alpha\in\mathbb R$. For any rational $p$ we have $\prod_k(pg_k(\alpha)+1)=(p\alpha+1)^n\ (*)$. Both the RHS and LHS of $(*)$ are polynomials which agree in infinitely many values of the argument $p$, so it must be that $g_k(\alpha)=\alpha,\ \forall k$. This is true for every real $\alpha$, so $f_k(\alpha)=f(1)\alpha,\ \forall k,\ \forall\alpha\in\mathbb R$, and we're done with the case $a\ne 0$. Now assume $a=0$, and that we can find $\alpha_i$ s.t. $f_i(\alpha_i)\ne 0$. Then, we can clearly find rationals $p_i$ s.t. for all $k$ we have $f_k(\sum_i p_i\alpha_i)=\sum_i p_if_k(\alpha_i)\ne 0$, which will contradict the fact that $\prod_k f_k$ evaluated in $\sum p_i\alpha_i$ is zero. This means that at least one of the $f_i$'s is identically zero (and for that $i$ we'll take $b=0$).