Inspired by Bulbuul Dev

Algebra Level 5

Let $i \ge 2, k \ge 2$ be positive integers and let $f(g\left( x \right) )$ be the number of distinct (not necessarily real) roots of the equation $g(x)=0$ .

Compare $\displaystyle A=f\left( \prod _{ n=1 }^{ i }{ \left({ x }^{ kn }-1\right) } \right)$ with $\displaystyle B=k\sum _{ n=1 }^{ i }{ n }$ .

Inspiration.

A=B

A < B

Not enough information Comparison Incorrect

A > B

1 solution

Aryaman Maithani
Jun 16, 2018

To evaluate A, we have to count the number of distinct solutions of:

$(x^k-1)(x^{2k}-1)\dots(x^{ik}-1)=0$

Each factor in itself has number of solutions equal to its degree (including complex solution)

$\therefore$ total number of solutions = $N = k + 2k + \dots ik = k\sum\limits_{ n=1 }^{ i }{ n } = B$

However, all the solutions counted above are not distinct, it can be clearly seen that all the roots of $(x^k-1)=0$ are also the roots of $(x^{2k}-1)=0$ .

$\therefore A < N$

$\implies \boxed{A < B}$

Inspired by Bulbuul Dev

1 solution

0 pending reports