It's not from set 1, it's from postal 2015 set 3.

We can prove that if \[a \omega^4 + b \omega^3 + c \omega^2 + d \omega = e, \text{\, \,where\,} a,b,c,d,e\in \mathbb Z^5\]then we must have \[a=b=c=d(=-e).\]We have \[(1+a_1 \omega)(1+a_2 \omega)(1+a_3 \omega)(1+a_4 \omega)(1+a_5 \omega)(1+a_6 \omega) = \left(\sum {a_1.a_2.a_3.a_4}\right)\omega^4+\left(\sum {a_1.a_2.a_3}\right)\omega^3 + \left(\sum {a_1.a_2}\right)\omega^2 + \left(\sum {a_1} +(a_1.a_2.a_3.a_4.a_5.a_6)\right)\omega + \sum {a_1.a_2.a_3.a_4.a_5} + 1.\]Therefore, since the above expression is an integer we must have \[\sum {a_1.a_2.a_3.a_4} = \sum {a_1.a_2.a_3} = \sum {a_1.a_2} = \sum {a_1} +(a_1.a_2.a_3.a_4.a_5.a_6).\]But we know that for positive integers $a_1,a_2,a_3,a_4,a_5,a_6,$ \[\sum {a_1.a_2.a_3} > \sum {a_1.a_2}\]Therefore, there are no such positive integers $a_1,a_2,a_3,a_4,a_5,a_6.$

Let us assume to the contrary that there exists such positive integers.Then $\omega$ is a root of the equation $a_1a_2a_3a_4a_5a_6x^6+....+\sum{a_i}x+1-k=0$.Then $\frac{1}{\omega}$ is also a root.However by Descates sign rule the equation has exactly 1 positive root.Contradiction!

JackXD wrote: Let us assume to the contrary that there exists such positive integers.Then $\omega$ is a root of the equation $a_1a_2a_3a_4a_5a_6x^6+....+\sum{a_i}x+1-k=0$.Then $\frac{1}{\omega}$ is also a root.However by Descates sign rule the equation has exactly 1 positive root.Contradiction! Why $\frac{1}{\omega}$ is also a root?

JackXD wrote: Let us assume to the contrary that there exists such positive integers.Then $\omega$ is a root of the equation $a_1a_2a_3a_4a_5a_6x^6+....+\sum{a_i}x+1-k=0$.Then $\frac{1}{\omega}$ is also a root.However by Descates sign rule the equation has exactly 1 positive root.Contradiction! Descarte's Rule of Signs can be applicable while talking about real roots. How would you define positive and negative for reals Satyaprakash2009rta wrote: JackXD wrote: Let us assume to the contrary that there exists such positive integers.Then $\omega$ is a root of the equation $a_1a_2a_3a_4a_5a_6x^6+....+\sum{a_i}x+1-k=0$.Then $\frac{1}{\omega}$ is also a root.However by Descates sign rule the equation has exactly 1 positive root.Contradiction! Why $\frac{1}{\omega}$ is also a root? Because complex roots exist in pairs.

utkarshgupta wrote: Satyaprakash2009rta wrote: Why $\frac{1}{\omega}$ is also a root? Because complex roots exist in pairs. Yeah I know, but the roots must be conjugate pairs. Are $\frac{1}{\omega}$ and ${\omega}$ conjugate pair??

Indeed, since $|\omega| = 1$, we have $\omega \cdot \overline{\omega} = 1$, so $1/\omega$ and $\omega$ are conjugates.