Given the following equation:
$$ x^{259}=1 $$
$$ x^{413}=1 $$
How many complex solutions for x have?
Thanks
Given the following equation:
$$ x^{259}=1 $$
$$ x^{413}=1 $$
How many complex solutions for x have?
Thanks
There are seven solutions of the form $e^{i2k\pi/7}$ for $k=0,...,6$, because the common factor of 259 and 413 is 7. Six of them are "complex" and the other one ($k=0$) is just $1$.
If you mean that $x$ must satisfy both equations, note that $\gcd(259,413)=7$.
Prove now: