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Good morning,

I have to resolve this system of equations: \begin{cases} x^2 \equiv 71 \pmod{7}\\ x^2 \equiv 71 \pmod{11}\\ \end{cases}

Then I find a part of the problem that I don't understand. Can someone explain me, how I find variable $x$ in a second degree congruent equation like this one: $$x^2 \equiv 1 \pmod{7} $$ or $$x^2 \equiv 5 \pmod{11}$$

Thank you

Joffan
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user3345696
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  • I don't understand it, so for example, the first one is asking to find a solution to $x^2\equiv 1\mod 7$ where $x\equiv \pm 1\mod 7$? – Daniel Aug 10 '17 at 08:12
  • The first one says: the solution to x2≡1mod7 is x≡±1mod7, but it doesn't explain the steps to find it. – user3345696 Aug 10 '17 at 08:22
  • Quasi describes how to start. Then you need to lift those solutions to solutions mod $77$. I describe how to do that e.g. here and here. – Bill Dubuque Aug 10 '17 at 15:42

1 Answers1

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\begin{align*} &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{For the first one:}\\[4pt] &x^2\equiv 1\;(\text{mod}\;7)\\[4pt] \iff\;&7\mid (x^2-1)\\[4pt] \iff\;&7\mid (x-1)(x+1)\\[4pt] \iff\;&7\mid (x-1)\;\;\text{or}\;\;7\mid (x+1)&&\text{[since $7$ is prime]}\\[4pt] \iff\;&x\equiv 1\;(\text{mod}\;7)\;\;\text{or}\;\;x\equiv -1\;(\text{mod}\;7)\\[4pt] \iff\;&x\equiv \pm 1\;(\text{mod}\;7)\\[8pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{For the second one:}\\[4pt] &x^2\equiv 5\;(\text{mod}\;{11})\\[4pt] \iff\;&x^2\equiv 16\;(\text{mod}\;{11})&&\text{[since $5 \equiv 16\;(\text{mod}\;{11})$]}\\[4pt] \iff\;&11\mid (x^2-16)\\[4pt] \iff\;&11\mid (x-4)(x+4)\\[4pt] \iff\;&11\mid (x-4)\;\;\text{or}\;\;11\mid (x+4)&&\text{[since $11$ is prime]}\\[4pt] \iff\;&x\equiv 4\;(\text{mod}\;{11})\;\;\text{or}\;\;x\equiv -4\;(\text{mod}\;{11})\\[4pt] \iff\;&x\equiv \pm 4\;(\text{mod}\;{11})\\[8pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Now the original question was to solve the system}\\[4pt] &x^2\equiv 71\;(\text{mod}\;7)\\[4pt] &x^2\equiv 71\;(\text{mod}\;11)\\[8pt] \text{But}\;\;& x^2\equiv 71\;(\text{mod}\;7)\\[4pt] \iff\;&x^2\equiv 1\;(\text{mod}\;7)\\[4pt] \iff\;&x\equiv \pm{1}\;(\text{mod}\;7)\\[8pt] \text{and}\;\;&x^2\equiv 71\;(\text{mod}\;{11})\\[4pt] \iff\;&x^2\equiv 5\;(\text{mod}\;{11})\\[4pt] \iff\;&x\equiv \pm{4}\;(\text{mod}\;{11})\\[8pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{So the original system is equivalent to the systems}\\[4pt] &x\equiv \pm{1}\;(\text{mod}\;7)\\[4pt] &x\equiv \pm{4}\;(\text{mod}\;{11})\\[2pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{The two equations above are actually $4$ distinct}\\[-1.25pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{systems, based on the choices for $\pm$. I'll do one of}\\[-1.25pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{the $4$ systems.}\\[8pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Thus, consider the system}\\[4pt] &x\equiv 1\;(\text{mod}\;7)\\[4pt] &x\equiv 4\;(\text{mod}\;{11})\\[2pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Equivalently,}\\[4pt] &x = 7a + 1,\;\text{for some $a \in \mathbb{Z}$}\\[4pt] &x = 11b + 4,\;\text{for some $b \in \mathbb{Z}$}\\[4pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{so then}\\[4pt] &7a + 1 = 11b + 4\\[4pt] \implies\;&11b = 7a -3\\[4pt] \implies\;&11b \equiv -3\;(\text{mod}\;7)\\[4pt] \implies\;&4b \equiv 4\;(\text{mod}\;7)\\[4pt] \implies\;&b \equiv 1\;(\text{mod}\;7)\\[4pt] \implies\;&b=7c+1,\;\text{for some $c \in \mathbb{Z}$}\\[4pt] \implies\;&x = 11(7c+1) + 4 = 77c + 15\\[4pt] \implies\;&x \equiv 15\;(\text{mod}\;{77})\\[8pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{One can easily verify that if}\\[4pt] &x \equiv 15\;(\text{mod}\;{77})\\[4pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{then the system}\\[4pt] &x\equiv 1\;(\text{mod}\;7)\\[4pt] &x\equiv 4\;(\text{mod}\;{11})\\[2pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{is satisfied. Thus we've solved one of}\\[-1.25pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{the $4$ systems. The others can be solved}\\[-1.25pt] &\!\!\!\!\!\!\!\!\!\!\!\!\! \text{in the same way.}\\[4pt] \end{align*}

quasi
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