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Let p be an odd prime number. Consider the set $S$ of CRS of $($mod $p^2)$. Now consider the subset $T$ $\in$ $S$ : $x_s$ $\equiv 2 ($mod $p)$.

What is the element $x_T$ of T that is NOT a primitive root $($mod $p^2)$? What is the process that can lead to find this one (or maybe many) solutions to the question?

The original problem was presented with p = 101

Thanks

Alessar
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  • By https://math.stackexchange.com/questions/227199/order-of-numbers-modulo-p2/229918#229918, either $2$ or $2+101$ is a primitive root $\pmod{101^2}$ – lab bhattacharjee Nov 26 '18 at 14:16

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You know that $2$ is a primitive root of $101^2$ and that $\phi(101^2) = 101\cdot 100.$ If one of the numbers $a$ in your set is not a primitive root then you must have $a^{100}\equiv 1 \pmod{101^2}.$ So you're looking for an element with order $100$.

You know $2^{101\cdot 100} \equiv 1 \pmod{101^2}.$ So $2^{101}$ has order dividing $100$. And by Fermat's little theorem, it's also congruent to $2 \pmod{101}$.

So the answer is $2^{101} \equiv 8385 \pmod{101^2}.$

B. Goddard
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