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# Help on these two questions please!

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1. Let $m$ be a positive integer, and suppose that $9$ is its own inverse $\pmod m$, but $3$ is $\textbf{not}$ its own inverse $\pmod m$. How many possible values for $m$ are there?

2. There are finitely many primes $p$ for which the congruence $$8x\equiv 1\pmod{p}$$has no solutions $x$. Determine the sum of all such $p$.

May 17, 2018

### 6+0 Answers

#1
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Heureka,  could you please tell me what

"9 is its own inverse" means

May 17, 2018
#2
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Hallo Melody.

In mathematics, in particular the area of number theory, a modular multiplicative inverse of an integer a is an integer x such that the product ax is congruent to 1 with respect to the modulus m. In the standard notation of modular arithmetic this congruence is written as

$$ax\equiv 1{\pmod {m}}$$

$$\text{We have a = 9 and x = the inverse of integer a } \\ 9x\equiv 1{\pmod {m}}}$$

So "9 is its own inverse" means that $$\text{ x=a=9 }$$

Finally

$$\begin{array}{|rcll|} \hline 9\cdot 9 &\equiv& 1 \pmod m \\ \hline \end{array}$$ heureka  May 17, 2018
edited by heureka  May 17, 2018
edited by heureka  May 17, 2018
edited by heureka  May 17, 2018
#3
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Thanks Heureka,

I understand about half of that,  at least I now know what 9 being its own inverse means.

So

9*9=1 (mod m)

81=1 (mod m)

so

80=0 (mod m)

Factors of 80 are     1, 2,4,5,8,10,16,20,40,80

This won't makes sense for 1,2,4,5, or 8

9*9 =81 = 1 (mod 10)

9*9 =81 = 1 (mod 16)

9*9 =81 = 1 (mod 20)

9*9 =81 = 1 (mod 40)

9*9 =81 = 1 (mod 80)

now 3*3=9 so this will stay 9 for any of the above mods.

So the possible values of m are  10,16,20,40, and 80       There are 5 of them

Is this right Heureka or am I not understanding something?

Melody  May 17, 2018
#4
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Hallo Melody,

this is right, but i think $$m =5$$ is a value for $$m$$ too.

Here my table:

$$\begin{array}{|r|r|r|r|r|} \hline m & & & =1 & \ne 1 \\ \hline 1 & 9*9=81\equiv 0 \pmod 1 & 3*3=9\equiv 0 \pmod 1 & & \checkmark \\ 2 & 9*9=81\equiv 1 \pmod 2 & 3*3=9\equiv 1 \pmod 2 & \checkmark & \\ 4 & 9*9=81\equiv 1 \pmod 4 & 3*3=9\equiv 1 \pmod 4 & \checkmark & \\ \color{red}5 & 9*9=81\equiv 1 \pmod 5 & 3*3=9\equiv 4 \pmod 5 & \checkmark & \checkmark \\ 8 & 9*9=81\equiv 1 \pmod 8 & 3*3=9\equiv 1 \pmod 8 & \checkmark & \\ \color{red}10 & 9*9=81\equiv 1 \pmod{ 10} & 3*3=9\equiv 9 \pmod{ 10} & \checkmark & \checkmark \\ \color{red}16 & 9*9=81\equiv 1 \pmod{ 16} & 3*3=9\equiv 9 \pmod{ 16} & \checkmark & \checkmark \\ \color{red}20 & 9*9=81\equiv 1 \pmod{ 20} & 3*3=9\equiv 9 \pmod{ 20} & \checkmark & \checkmark \\ \color{red}40 & 9*9=81\equiv 1 \pmod{ 40} & 3*3=9\equiv 9 \pmod{ 40} & \checkmark & \checkmark \\ \color{red}80 & 9*9=81\equiv 1 \pmod{ 80} & 3*3=9\equiv 9 \pmod{ 80} & \checkmark & \checkmark \\ \hline \end{array}$$

So the possible values of m are  5,10,16,20,40, and 80       There are 6 of them heureka  May 17, 2018
#5
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Yes of course, I missed that one. Thanks Heureka :))

Melody  May 17, 2018
#6
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Thanks everybody so much

May 17, 2018