2

$x^2 + 3x +1 \equiv 0 \pmod{101}$. To solve this I found the determinant $D = 5 \pmod{101}$). Using the Legendre symbol,

$$\left(\frac{5}{101}\right) = \left(\frac{101}{5}\right) \equiv \left(\frac{1}{5}\right) \equiv 1,$$

$\therefore$ The equations have a solution.

My question is how I can find out how many solutions it has?

Bob Happ
  • 642
  • 1
  • 6
  • 24
Orland
  • 156
  • 8

4 Answers4

2

Since $$ (x-49)^2-77\equiv x^2+3x+1\pmod{101} $$ we are looking for solutions to $(x-49)^2\equiv77\pmod{101}$. You have verified that there is a root, so $77$ is a quadratic residue mod $101$, thus, there are two solutions for $$ (x-49)^2\equiv77\pmod{101} $$ Alternatively, working mod $101$, by squaring and multiplying $$ 77^2\equiv71\\ 77^3\equiv13\\ 77^6\equiv68\\ 77^{12}\equiv79\\ 77^{24}\equiv80\\ 77^{25}\equiv100\\ 77^{50}\equiv1 $$ Therefore, $77$ is a quadratic residue mod $101$. Thus, there are two solutions to $x^2+3x+1\pmod{101}$.

robjohn
  • 355,927
1

An idea to tackle this and similar questions using as small numbers and multiples of $\;101\;$ as possible (when possible, of course...).

Observe that $\;5=-96\pmod {101}\;$ , and $\;96=2^5\cdot3\;$ , so we can try to deal with these apparently easier numbers. Since we have $\;96=16\cdot6\;$, we have:

$$4^2=16=2^4\;,\;\;101\cdot2=202-6=196=14^2$$

and thus we have:

$$\;\sqrt{16}=4\;,\;\;\sqrt{-6}=14\implies \sqrt{16\cdot(-6)=-96}=4\cdot14=56$$

and thus also $\;\sqrt{-96}=-56=45\;$ , so $\;x^2=5\pmod{101}\implies x=\pm56=56,\,45\;$

Finally, your quadratic's solutions are ($\pmod{101}$ ):

$$x_{1,2}=\frac{-3\pm56}2=\begin{cases}\frac{98+56}2=77\\{}\\\frac{98-56}2=21\end{cases}$$

DonAntonio
  • 215,504
0

If the discriminant of a quadratic is a nonzero square modulo odd prime $p$, then the quadratic has exactly two roots mod $p$.

Robert Israel
  • 473,652
  • Okay, but with what I found with the Legendre Symbol, how can I find out that there is two roots. – Orland Dec 07 '16 at 05:59
  • 1
    @Orland: The quadratic formula tells you that the solutions are $$x=\frac{-3\pm\sqrt{D}}2.$$ There are no solutions in a domain of interest, if $D$ has no square root in that domain. There are two solutions if $D$ has a non-zero square root, and one double root if $\sqrt{D}=0$. The reasoning we all learned in school, relevant to the field of real numbers, applies to all fields where $2\neq0$. Legendre symbol tells you that there is a $\sqrt{5}\in\Bbb{Z}_{101}$. – Jyrki Lahtonen Dec 07 '16 at 07:09
0

For prime $p$: If $A$ is a square modulo $p,$ take any B such that $A\equiv B^2\pmod p.$ Then $$x^2\equiv A \pmod p\iff (x-B)(x+B)\equiv x^2-B^2\equiv 0 \pmod p\iff$$ $$\iff (\;p|(x-B)\lor p|(x+B)\;)\iff x\equiv \pm B \pmod p.$$ If $A\not \equiv 0 \pmod p,$ then $B\not \equiv 0\pmod p.$ If $p\ne 2,$ and $B\not \equiv 0 \pmod p$ then $B\not \equiv -B \pmod p.$ So for $p\ne 2$ and $A\not \equiv 0\pmod p$ there are, modulo $p,$ exactly $2$ solutions to $x^2\equiv A \pmod p,$ which are $x\equiv \pm B \pmod p.$

In your Q you are looking for solutions to $(2x+3)^2\equiv 5 \pmod {101}.$ Take any $B$ such that $B^2\equiv 5 \pmod {101}.$ Then $(2x+3)^2\equiv B^2 \iff 2x+3\equiv \pm B \pmod {101},$ which implies there exactly $2$ values, modulo $101$, for $x$.

DanielWainfleet
  • 60,123