1

Quite probably a silly question, but if I have:
$a^{2} \equiv 1 \pmod n$, is this the same as $a \equiv 1 \pmod n$?

Tom E.
  • 101
  • No. The number of solutions depends on the number of prime factors of $n$. Your conclusion is valid only when $n=2$. Otherwise we have at least $a\equiv\pm1$ as possibilities. With $n=3\cdot5$ we have $a\equiv\pm1$ and $a\equiv\pm4$ because $4$ is congruent to $\pm1$ modulo both $3$ and $5$. You can similarly ChineseRemainderTheoremCombine $\pm1$ modulo all the prime power factors of $n$. – Jyrki Lahtonen Sep 07 '17 at 08:16
  • See here for another example with four solutions. – Jyrki Lahtonen Sep 07 '17 at 08:22
  • 1
    Thank you @JyrkiLahtonen. So it would be valid to say $ a \equiv 1 \pmod n$ is one possible solution, if I'm not necessarily looking to find all the solutions, but only a specific one for whatever I'm doing. – Tom E. Sep 07 '17 at 08:29
  • 1
    Yes, Tom. That is correct. If $a\equiv1\pmod n$ then also $a^2\equiv1\pmod n$, so that is one possible solution. – Jyrki Lahtonen Sep 07 '17 at 08:31

0 Answers0