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Consider the Group U(55) under multiplication modulo 55. Let x∈U(55) be such that x.x=26 and x>30. Then find the value of x.

Now I want to know that if the above question can be represented in congruence equation. If yes then please tell me how it would be done and recommend a book or something from which I can learn and practice such questions.

Saransh Chauhan
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  • $x^2\equiv 26\pmod {55}$. To solve it I would suggest noting that $55=5\times 11$ so you can solve $\pmod 5$ and $\pmod 11$ separately. Alternatively, $55$ is very small so trial and error works just fine. – lulu Apr 04 '20 at 13:20
  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Apr 04 '20 at 13:22

1 Answers1

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You basically want to find $x>30$ such that $x^2\equiv26\pmod{55}$.

This is true if and only if $x^2\equiv 26\equiv1\pmod{5}$ and $x^2\equiv 26\equiv4\pmod{11}$.

It is easy to check that the solutions to these congruences larger than 30 are $x=31,46$.

Here is how to solve the congruences. First notice that $x^2\equiv1\pmod 5$ iff $x\equiv\pm 1\pmod5$, and $x^2\equiv4\pmod{11}$ iff $x\equiv\pm2\pmod{11}$.$^{\color{red}1}$ Thus we have four possibilities: \begin{align*} x&\equiv 1 \pmod 5 &\text{and}&& x &\equiv 2 \pmod {11}\\ x&\equiv 1 \pmod 5 &\text{and}&& x &\equiv -2 \pmod {11}\\ x&\equiv -1 \pmod 5 &\text{and}&& x &\equiv 2 \pmod {11}\\ x&\equiv -1 \pmod 5 &\text{and}&& x &\equiv -2 \pmod {11} \end{align*}

In general, to solve $x\equiv a_1\pmod{M_1}$ and $x\equiv a_2\pmod{M_2}$, observe that $$x=a_1M_2y_2+a_2M_1y_1$$

is a solution, where $y_2={M_2}^{-1}\pmod{M_1}$ and $y_1={M_1}^{-1}\pmod{M_2}$. Indeed, if you work $\bmod M_1$, the second term vanishes and the first term simplifies to $a_1$, whereas $\bmod M_2$, the first term vanishes and you're left with $a_2$.$^{\color{red}2}$

Reasoning this way, we get the four solutions \begin{align*} x&=1\cdot 11 \cdot 11 + 2\cdot5 \cdot 9 &{}={}& 211\equiv46\pmod{55}\\ x&=1\cdot 11 \cdot 11 + (-2)\cdot5 \cdot 9 &{}={}& 31\\ x&=(-1)\cdot 11 \cdot 11 + 2\cdot5 \cdot 9 &{}={}& -31\equiv24\pmod{55}\\ x&=(-1)\cdot 11 \cdot 11 + (-2)\cdot5 \cdot 9 &{}={}& -211 \equiv 9 \pmod{55}, \end{align*} two of which are $>30$.

There are loads of resources available online, a quick DuckDuckGo search gave this set of notes on elementary number theory which covers this sort of thing.

$^1$ In general, there are always 2 or 0 solutions to $x^2\equiv a\pmod m$. Read about quadratic residues for more information.

$^2$ This is essentially how you prove the Chinese remainder theorem. Also, finding only one solutions is enough, because linear congruences always have one solution (up to modulo).

Luke Collins
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