Part BCSIR NET December 2024a-square-is-never-a-primitive-root

A square is never a primitive root

Let p be a prime number. An element a of the multiplicative group $(Z / p Z)$ ˣ is said to be a primitive root in $(Z / p Z)$ ˣ if the order of a in $(Z / p Z)$ ˣ is p − 1. Let S_p be the number of primitive roots in $(Z / p Z)$ ˣ and $φ$ denote the Euler $φ -$ function. Which of the following statements is true?

A.For each $p < 100, \sum (n = 1..\infty) (S_{p} / φ (p))^{n}$ converges.
B.For each $100 \leq p \leq 200, \sum (n = 1..\infty) (S_{p} / φ (p))^{n}$ diverges.
C.For each $p > 200, \sum (n = 1..\infty) (S_{p} / φ (p))^{n}$ converges.✓
D.The element 4 mod 101 in $(Z /101 Z)$ ˣ is a primitive root.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: If d divides |G| then G has a subgroup of order d

A square is never a primitive root

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