Proof for a relation between prime numbers and roots of unity

Question

Property:

if $n = p+q$, $\;$$p$ and $q$ are primes then $p$th and $q$th roots of unity are also the $n$th roots of unity.

Example 1: The 4th roots of unity are $\pm1,\;\pm i$.

As 4 = 2+2 and the 2nd roots of unity are $\pm 1$ and these are also 4th roots of unity.

Example 2: Solutions for $x^6 = 1$ are: $ \\ x_1=1\\ x_2=−1\\ x_3=0.5+0.86603i\\ x_4=0.5−0.86603i\\ x_5=−0.5+0.86603i\\ x_6=−0.5−0.86603i\\ $

As 6= 2+2+2

The 2nd roots of unity are $\pm1$ which are also a part of 6th roots of unity

and 6 = 3+3

The 3rd roots of unity are: $ x_1=1\\ x_2=−0.5+0.86603i\\ x_3=−0.5-0.86603i\\ $

These are also 6th roots of unity.

Can anyone show the proof of this identity using elementary(high school level) maths?

Can I prove this by induction or contradiction, or do I have to use representing complex numbers in polar or exponential form for proving this?

My approach was using the angles between the roots of unity but I have no progress.

---

Sorry for long question

Answer 1

Conterexample to your literal claim: $5=2+3$, but the fifth roots of unity are neither 2nd nor 3rd roots of unity (except $1$).

But if $n=kp$ then every solution of $X^p=1$ is also a solution of $X^n=1$: Let $\zeta\in \Bbb C$ with $\zeta^p=1$. Then $$\zeta^n=\zeta^{pk}=(\zeta^p)^k=1^k=1.$$ For this observation, $p$ need not be prime.

Answer 2

It is not true in general. Let $n=8=5+3$ and $\omega\neq 1$ be a cube root of unity. Then $\omega^8=\omega^2\neq 1$.

It is true that if $d\mid n$, then any $d^{{\rm th}}$ root of unity is also an $n^{{\rm th}}$ root of unity.