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I am looking for a ring $R$ in which $2$ is invertible and there are four solutions to $x^2-1=0$.

$R=\Bbb Z/8\Bbb Z$ has the four solutions $1,3,5,7$ to $x^2-1=0$, but $2$ is not invertible.

MyNameIs
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  • See here, here or ... – Jyrki Lahtonen Jun 14 '16 at 11:14
  • You should "make maths" adding "....there are EXACTLY four solutions..." because strictly speaking, all ring having plus of four solutions is an answer to your question. For example, the quaternions have a non countable infinity of solutions to $x^2-1=0$ so have as well four solutions. – Piquito Jun 14 '16 at 12:31

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The ring $R=\Bbb{Z}/15\Bbb{Z}$ seems to fit the bill. The solutions are (the cosets of) $\pm1$ and $\pm4$. A process finding the solutions amounts to finding the solutions modulo $3$ and $5$ (two for each), and then combining them using the Chinese Remainder Theorem. There are many examples of how to do that on our site.

Jyrki Lahtonen
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  • In general this equation has exactly four solutions in $\Bbb{Z}_m$ when the odd integer $m$ has exactly two prime factors. – Jyrki Lahtonen Jun 14 '16 at 10:54
  • Could you elaborate on how you used the Chinese remainder theorem? It gives us an isomorphism $$Z_{15}\cong Z_5\times Z_3$$ how do you use that? – MyNameIs Jun 14 '16 at 10:55
  • @MyNameIs: If $f:\Bbb{Z}_{15}\to\Bbb{Z}_5\times\Bbb{Z}_3$ is the ring isomorphism, then the four solutions are $f^{-1}(\pm1,\pm1)$. All four sign combinations. – Jyrki Lahtonen Jun 14 '16 at 10:58
  • It is easy to see that if $p$ is an odd prime, the only solutions in $\Bbb{Z}{p^a}$ are $\pm1$. The same idea then finds the four solutions in $\Bbb{Z}{p^aq^b}$, $p,q$ distinct odd primes. – Jyrki Lahtonen Jun 14 '16 at 11:00
  • Clever, so you were just looking for an $n$ with $2$ invertible in $\Bbb Z_n$ which can be written as a product of two primes? – MyNameIs Jun 14 '16 at 11:01
  • That's correct. Any product of two powers of odd primes will work. If you use the product of three powers of odd primes you get eight solutions by the obvious recipe. – Jyrki Lahtonen Jun 14 '16 at 11:02
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Note that this equation is the same as $$ (x - 1)(x + 1) = 0 $$ so, you're looking for a ring in which there are two zero-divisors $a,b$ such that $b - a = 2$ (so that $a = x-1$ and $b = x+1$) and $ab = 0$.

If we restrict our search to $\Bbb Z / n\Bbb Z $, then the only non-invertible elements are the factors of $n$ besides $1$, which are also zero-divisors. So, $n$ will be satisfactory if and only if:

  • $2$ does not divide $n$
  • $n \mid a(a + 2)$ for some $a$ for which $n \not \mid a$ and $n \not \mid (a+2)$.

So, for example: $15 = 3 \times 5$ will work, as will $35 = 5 \times 7$, as will $21$ which divides $7 \times 9$.

Ben Grossmann
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