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 $(\mathbb{Z}/p\mathbb{Z})$ ˣ is said to be a primitive root in $(\mathbb{Z}/p\mathbb{Z})$ ˣ if the order of a in $(\mathbb{Z}/p\mathbb{Z})$ ˣ is p − 1. Let S_p be the number of primitive roots in $(\mathbb{Z}/p\mathbb{Z})$ ˣ and $\varphi$ denote the Euler $\varphi-$ function. Which of the following statements is true?

A.For each $p < 100, \sum(n=1..\infty) (S_p/\varphi(p))^{n}$ converges.
B.For each $100 \le p \le 200, \sum(n=1..\infty) (S_p/\varphi(p))^{n}$ diverges.
C.For each $p > 200, \sum(n=1..\infty) (S_p/\varphi(p))^{n}$ converges.✓ Correct
D.The element 4 mod 101 in $(\mathbb{Z}/101\mathbb{Z})$ 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, and why this one tests standard counterexample.

See pricing

50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

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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