## Position of roots

The following result is quite useful in irreductibility problems for polynomials.

Prove that if then all the roots of satisfy or a root of unity of some order.

**Solution:**

If all are equal, then the polynomial is a scalar multiplied with , and all roots satisfy . In the following, suppose there exist two numbers which are not equal.

Suppose and (obviously since ). Then we have . Multiplying this last relation with which is not zero, since 1 is not a root of , we get . Taking modules and using the modulus inequality, we get , so we get the equality. Therefore, any two non-zero terms are positive real scalar multiple of each other. Since there exists such that and , we can find such that which yelds . Therefore, a root can be of modulus strictly greater than 1, or a positive scalar multiple of a root of unity. Since assuming we get a contradiction ( we couldn’t have equality in the triangle inequality ), we see that if then is a root of unity of some order.