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I need to understand the concept behind a non trivial square root. Also how to answer these two questions and how to get to the answer?

  1. Give a non-trivial square root of 30

  2. Give a non-trivial square root in the integers for (mod 143)

Bill Dubuque
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Bob
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    Add some context, please! What's the ring in Q1? What are you trying to find a square root of in Q2? I might hazard a guess that non-trivial square roots emerge when a residue class has multiple square roots in the residue class ring. E.g. $\pm1$ and $\pm4$ are all square roots of $1$ in the ring $\Bbb{Z}_{15}$. Supposedly $\pm4$ could be called non-trivial, but this is just a guess. Chinese remainder theorem is your friend here, if this guess is correct. – Jyrki Lahtonen Apr 29 '14 at 23:08
  • I have not been given any more information just been asked to give a non trivial square root of 30 and again same thing for q2 :s – Bob Apr 29 '14 at 23:13
  • Then I would guess that you are expected to cook up the rest. For Q1 find a modulus such that 30 has four distinct modular square roots, and for Q2 find a residue class that has non-trivial square roots (hint: what's square root of $1$ modulo 143?). I have not heard of the concept of a non-trivial square root before, so it is possible that the term is private to your teacher, in which case I cannot help. As you tagged this with primality-test, I am somewhat optimistic about having guessed correctly. Not betting on it though. – Jyrki Lahtonen Apr 30 '14 at 00:09
  • @Jyrki There are plenty of mentions of nontrivial square roots here. – Bill Dubuque Apr 30 '14 at 01:45

1 Answers1

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A nontrivial square root of $a^2$ means a root $\,b \ne \pm a,\,$ so the quadratic $\,x^2 - a^2 = (x-a)(x+a)\,$ has $> 2\,$ roots $\,x = \pm a, b.\, $ Then $\,(b-a)(b+a) = 0\,$ but $\,b\pm a \ne 0,\,$ so $\,b\pm a\,$ is a zero divisor.

Generally we can quickly factor $\,n\,$ given any polynomial which has more roots mod $\,n\,$ than its degree, so any nontrivial idempotent or nontrivial square-root will split $\,n,\,$ since it yields a quadratic with $3$ roots. See this answer for a hint on how to find nontrivial square roots mod composite $\,n.\,$ There we find nontrivial idempotents, but the same idea works for square roots.

Bill Dubuque
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  • I didn't understand why a quadratic equation should have more than 2 roots. Would you mind letting me have an example (and please explain it in a simple way so I may understand; doesn't matter if you include a bit of modular arithmetic and number theory)? – Spectre Jul 21 '21 at 05:53
  • @Spectre Not "should" but "could", e.g. $,\bmod 8!:\ x^2\equiv 1,$ for $,\pm x\equiv 1,3,,$ i.e. $,x\equiv 1,3,5,7,,$ i.e. $,{\rm odd}^2\equiv 1\pmod{!8}\ \ $ – Bill Dubuque Jul 21 '21 at 10:08
  • so basically it's something pertaining to modular arithmetic? – Spectre Jul 22 '21 at 08:31