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I wonder about an equation given as a solution to the following task:

Calculate the multiplicative inverse $$5^{−1} \pmod {13}$$

The solution ends in this equation: $$ 5^{11} \equiv 5^{10} \cdot 5 \equiv −1\cdot 5 \equiv 8 \pmod {13} $$ and the check:

$$ 5 \cdot 8 \equiv 40 \equiv 1 \pmod{13} $$

1) What is the inverse here now?

2) Why is $5^{10} \equiv -1 \pmod {13}$?

As far as I understood Fermat the rule is given as: $$ x^{(p−1)} \equiv 1 \pmod p $$ which won't come into account here, since in this case $p = 13$ and not ideal for calculating $5^{11}$.

If I would use $x^{p−1}$ it would be $5^{12}$ in this case, so I'd still need to divide by $5^1$ to get to the result of $5^{11}$.

I am so confused although it should be really easy.

Bill Dubuque
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yesyoor
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  • Note $5^2\equiv-1\pmod{13}$ – J. W. Tanner Jan 19 '20 at 22:13
  • Another hint : $5^{10}=\bigl(5^2\bigr)^5$. – Bernard Jan 19 '20 at 22:16
  • an extension of Fermat says : $a^{(p-1)k+b}\equiv a^b\pmod p$ as long as $a$ is not a multiple of $p$. By that we get $a^{(p-1)1+(-1)}\equiv a^{-1}\pmod p$ –  Jan 19 '20 at 22:18
  • @PrudiiArca no they want me to calculate the multiple inverse of 5^(-1) mod 13 – yesyoor Jan 19 '20 at 22:40
  • thanks @J.W.Tanner and Bernard now I finally got it – yesyoor Jan 19 '20 at 22:44
  • @PrudiiArca ok you might be right, I am not a native english speaker so I might have made a mistake translating it – yesyoor Jan 20 '20 at 00:03
  • You need to tell us what is written before the end of the solution if we are to have any hope of explaining what the author intended. That's very hard to guess given only the small excerpt you gave, i.e. it is not at all clear how the author computed $,5^{10}\equiv -1,$ from what is written in the question. – Bill Dubuque Jan 20 '20 at 00:14
  • at yesyoor no worries. @BillDubuque alright – Jonas Linssen Jan 20 '20 at 07:59

2 Answers2

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Question 2: Why is $5^{10}\equiv - \pmod {13}$.

Well....Because it is.

$5^2 \equiv 25\equiv -1\pmod {13}$ so $5^{10}\equiv (-1)^5\equiv -1 \pmod {13}$.

The real question is why are we trying to find $5^{10}$?

$5^{11}*5 \equiv 5^{12} \equiv 1 \pmod {13}$ so $5^{-1}\equiv 5^{11}$ and..... well, I guess the author just figured $5^{10}$ could be easily calculated by subsequent squaring. $5^{11} \equiv 5*5^{10}\equiv 5*(5^2)^5$ is an easy way do to it.

We could, just to be different figure $5^{11} = (5^3)^3*5^2\equiv 125^3*(-1)\equiv -8^3\equiv (-8)^3 \equiv 5^3 \equiv 125 \equiv 8\pmod {13}$ but that would be unnecessarily complicated.

Or we could just do increasing powers and see when we return to $1$. $5^1 \equiv 5$ and $5^2 \equiv -1$ so $5^3\equiv -1*5 \equiv 8$ and $5^4\equiv 1$. So $5^{4} =1$ and $5^{-1} \equiv 5^3 \equiv 8$

To be honest.... I do not know why the author assumed that it'd be "obvious" that $5^{10} \equiv -1$ but not that it'd be "obvious" that $5^{11}\equiv 8$.

But one way or another..... $5^{11}*5 \equiv 1\pmod {13}$ so $5^{-1}$ is $5^{11}$ whatever that is. And roll up your sleeves and do it... $5^{11}\equiv 8\pmod {13}$ by one method or another.

Question 1:

What is the inverse? Well, since $5*8\equiv 1$ it is $8$

fleablood
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  • Simpler: $\bmod 13!:,\ 5(5)\equiv -1,\Rightarrow, \color{#c00}{-5}(5)\equiv 1,\Rightarrow, 5^{-1}\equiv \color{#c00}{-5}\equiv 8\ \ \ $ – Bill Dubuque Jan 20 '20 at 00:22
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Fermat gives you $5^{12} \equiv 1 \operatorname{mod} 13$. But to have a multiplicative inverse to $5$ you have to multiply it with $5$ to get the result. Hence you have to write $5 \cdot 5^{11} \equiv 1 \operatorname{mod} 13$. The inverse thus is given by $5^{11} \equiv 8 \operatorname{mod} 13$.

Jonas Linssen
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