# 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.

• What can you prove, and how have you proved it? Please add all the details in your post. Commented Oct 16, 2022 at 17:03
• I've updated my original post with a proof outline for the case of 5 $a_j$'s. Commented Oct 16, 2022 at 21:28
• Side note: The five-term case can also be solved by noting that $1+\omega a_j$ has argument strictly between $\frac\pi5$ and $\frac{2\pi}5$ (with the same conclusion that the product lies in the lower half-pane). Commented Oct 16, 2022 at 23:31

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 $$S_2=\sum_{1\le i $$=\sum_{1\le i $$\le \left(\sum_{1\le i$$\le \left(\sum_{1\le i $$<\sum_{1\le i contradicting $$S_2=S_3.$$
• @JeremiahGoertz . I subtracted polynomial multiples of $G(x)$ from $F(x)-L$ to get rid of the terms in $x$ of degree $>4.$ I expected to get something useful because $\omega$ is a zero of the resulting 4th-degree polynomial, and $\omega$ is already known to be a zero of another one ($G(x)$). I could have also eliminated the term in $x^4$ but when I looked at $H(x)$ I had a different idea. Commented Oct 18, 2022 at 3:12