A number-theoretic problem involving 5th roots of unity This problem is supposed to be from an Indian Math Contest, although I couldn't find any reference to it online.  To wit:
Let $\omega = e^{2\pi i/5}$.  Prove that there are no positive integers $a_1, a_2, a_3, a_4, a_5, a_6$ such that
$(1 + \omega a_1)(1 + \omega a_2)(1 + \omega a_3)(1 + \omega a_4)(1 + \omega a_5)(1 + \omega a_6)$
is an integer.
I can prove the related statement for 5 $a_j$'s, but a proof of the given statement eludes me.
Edit:  The basic idea behind the proof for the case of 5 $a_j$'s is to expand the product as a polynomial in $\omega$.  Then note that the coefficient for $\omega^3$ is greater than that of $\omega^2$, and the coefficient of $\omega^4$ is greater than that of $\omega^1$.  This forces the product
$(1 + \omega a_1)(1 + \omega a_2)(1 + \omega a_3)(1 + \omega a_4)(1 + \omega a_5)$
to have a negative imaginary part, and thus it cannot be an integer.
 A: Let $F(x)=\prod_{j=1}^6(1+x a_j)=1+\sum_{j=1}^6x^jS_j$ and suppose $F(\omega)=L\in\Bbb Z.$
Let $G(x)=\sum_{j=0}^4x^j.$ We have $G(\omega)=0.$
So $0=H(\omega)$ where $$H(x)=F(x)-x^2S_6G(x)-x(S_5-S_6)G(x)-L=$$ $$=x^4(S_4-S_5)+x^3(S_3-S_5)+x^2(S_2-S_5)+x(S_1+S_6-S_5)+(1-L).$$ Now $G(x)$ is irreducible in $\Bbb Z[x]$ because the substitution $x=y+1$ into $G(x)$ yields a polynomial in $y$ that satisfies Eisenstein's Criterion. So if $P(x)\in\Bbb Z[x]$ and $P(\omega)=0$ then $P(x)$ is divisible in $\Bbb Z[x]$ by $G(x).$ In particular, when $P(x)=H(x),$ we have $H(x)=kG(x)$ for some $k\in \Bbb Z.$  So the co-efficients of $x^2$ and $x^3$ in $H(x)$ are each equal to $k$, which implies $$S_2=S_3.$$ Now WLOG suppose $a_6=\min \{a_1,...,a_6\}.$ We have $$S_2=\sum_{1\le i<j\le 6}a_ia_j=$$ $$=\sum_{1\le i<j\le 5}a_ia_j+\sum_{j=1}^5a_ja_6\le$$ $$\le \left(\sum_{1\le i<j\le 5}a_ia_ja_6\right)+(a_1a_2+a_2a_3+a_3a_4+a_4a_5 +a_5a_1)\le$$$$\le \left(\sum_{1\le i<j\le 5}a_ia_ja_6\right)+(a_1a_2a_3+a_2a_3a_4+a_3a_4a_5+a_4a_5a_1 +a_5a_1a_2)<$$ $$<\sum_{1\le i<j<k\le 6}a_ia_ja_k=S_3$$  contradicting $S_2=S_3.$
