Another method to show the "only if " direction is to use the fact that the trace of $\zeta_n$ is equal to zero if n is not square free, while by definition, the trace of $\zeta_n$ in this case is exactly the same as the sum of all the primitive n-th roots of unity, so we have a linearly dependent relation over $\mathbb {Q}$ for all the primitive n-th roots, so they could not form a basis, see ...
Checking per Gerry's suggestion, a quick spreadsheet for the 40 primitive roots mod 101 shows that twenty-six (26) of them are square-free and fourteen (14) of them are not. We are helped in this by the fact that 2 is the smallest primitive root mod 101, so taking powers of 2 with exponents coprime to 100 gives all forty of the primitive roots (reduced mod 101).
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How would you find a primitive root of a prime number such as 761? How do you pick the primitive roots to test? Randomly? Thanks
Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple of 3, and c cannot be a multiple of 3. My attempt: Let a and b be relatively prime positive in...
The definition, given in the text, of primitive central idempotent element $e$ is if $e$ is central and has no proper decomposition as a sum of orthogonal central idempotent elements.
This is because all odd squares are $1\pmod 8$. Any primitive triple must have two odd squares, whose difference is therefore a multiple of $8$, and so the third number must be a multiple of $4$. (You can't have two odd numbers with even hypotenuse, since the sum of two odd squares is $2\pmod 4$, but every even square is $0\pmod 4$.)