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Solve $ 2^x \equiv x \pmod {11}$.

I know 2 is a primitive root modulo 11.

So. I get $x \equiv \operatorname{ind}_2x \pmod {10}$

And I'm stuck!

(Maybe I can $x=1$, $x=2$, $x=3$, and so on... but looking for brilliant method.)

Could you give me some advice?

Is 'primitive root' useful thing for solving this type?

Thanks in advance.

$$ $$ P.S.

And I'd like to check this via http://wolframalpha.com

but this show.... http://www.wolframalpha.com/input/?i=2%5Ex++mod11+%3Dx

I wanna know proper command for check this problem.

maybe not [ 2^x mod11 =x ]

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    $\begingroup$ I suggest trial and error $\endgroup$ – Hagen von Eitzen Nov 28 '14 at 7:39
  • $\begingroup$ @HagenvonEitzen Thank you for your comment. And you mean substitution for x? $\endgroup$ – user143993 Nov 28 '14 at 7:41
  • $\begingroup$ I think you mean $x\equiv \text{ind}_2(x)$? $\endgroup$ – David Peterson Nov 28 '14 at 7:43
  • $\begingroup$ @DavidPeterson Really thank you. I edited :-) $\endgroup$ – user143993 Nov 28 '14 at 7:48
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    $\begingroup$ Wolfram $\endgroup$ – Quang Hoang Nov 28 '14 at 8:02
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Using Fermats little theorem, it follows that $2^x\equiv2^{x \mod 10}$. This should be equal to $x\mod11$. The residue classes are from $x\equiv0$ to $x\equiv9$: $$1;2;4;8;5;10;9;7;3;6;$$

This should be equal to the residue class $x\mod11$. Using the Chinese remainder theorem: \begin{align} x\equiv0\mod 10\qquad &x\equiv1&\mod11\implies &x\equiv100&\mod110\\ x\equiv1\mod 10\qquad &x\equiv2&\mod11\implies &x\equiv101&\mod110\\ x\equiv2\mod 10\qquad &x\equiv4&\mod11\implies &x\equiv92&\mod110\\ x\equiv3\mod 10\qquad &x\equiv8&\mod11\implies &x\equiv63&\mod110\\ x\equiv4\mod 10\qquad &x\equiv5&\mod11\implies &x\equiv104&\mod110\\ x\equiv5\mod 10\qquad &x\equiv10&\mod11\implies &x\equiv65&\mod110\\ x\equiv6\mod 10\qquad &x\equiv9&\mod11\implies &x\equiv86&\mod110\\ x\equiv7\mod 10\qquad &x\equiv7&\mod11\implies &x\equiv7&\mod110\\ x\equiv8\mod 10\qquad &x\equiv3&\mod11\implies &x\equiv58&\mod110\\ x\equiv9\mod 10\qquad &x\equiv6&\mod11\implies &x\equiv39&\mod110\\ \end{align}

The given residue classes $\mod110$ are the answer to the question.

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    $\begingroup$ I think everyone else who looked at this question assumed it would only depend on $x$ modulo 11, including myself. Great work! $\endgroup$ – RghtHndSd Nov 28 '14 at 16:45
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    $\begingroup$ Don't you mean mod $110$ instead of $121$? $\endgroup$ – Barry Cipra Nov 28 '14 at 17:03
  • $\begingroup$ Yes, thanks for pointing this out, @Greg thanks for editing. $\endgroup$ – Thijs Nov 28 '14 at 17:45
  • $\begingroup$ Wow.. Really great!! wow....... wow.. both modulo 10 and 11...!!! Thank you for your answer $\endgroup$ – user143993 Nov 30 '14 at 2:54
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Let $a$ be any nonnegative integer and let $k=a-2^a$. Let $x=a+10k$. Then, using a mixture of unconditional equality and mod-$11$ congruence, we have

$$2^x=2^{a+10k}\equiv2^a=a-(a-2^a)=a-k\equiv a+10k=x$$

This doesn't give every solution, but it does give an infinite family of solutions (all negative). For example, $a=1$ gives $2^{-9}\equiv-9$ mod $11$.

Added later: In light of Thijs's complete solution, it's worth noting that the explicit formula here, which can be written as

$$x=11a-10\cdot2^a$$

gives, for $a=0,1,2,\ldots,9$, the values $-10,-9,-18,\ldots,-5021$, which agree mod $110$ with the residues in Thijs's Chinese remainder theorem results, in the exact same order, i.e., $-10\equiv100$, $-9\equiv101$, $-18\equiv92,\ldots,-5021\equiv39$ mod $110$. Presumably this generalizes in some fashion to something like

$$x\equiv q^x\text{ mod } p\iff x\equiv pa-(p-1)q^a\text{ mod }p(p-1)\text{ for some }a\in\mathbb{N}$$

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This is Mathematica code. I believe that it will work on Wolfram Alpha.

m = 11;
Select[Range[m], Mod[2^#, m] == # &]
{7}

For large values of $m$ this will be faster

m = 1234567;
Select[Range[m], PowerMod[2, #, m] == # &]
{313692}

After reading the comments and Thijx answer, I propose the following code:

m = 11;
xmax = EulerPhi[m]m;
Select[Range[xmax], Mod[2^# - #, m] == 0 &]
{7, 39, 58, 63, 65, 86, 92, 100, 101, 104}
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  • $\begingroup$ Can you explain the code please? $\endgroup$ – Mmm Nov 28 '14 at 16:11
  • $\begingroup$ I acknowledge it is quite cryptical. Range[m] is $\{1,\dots,m\}$. Mod[2^a,b]=PowerMod[2,a,b] computes $2^a\mod b$. The #, & construction is a pure function. # is the argument. It returns True if the equality holds, False if not. Select selects the values in Range[m] returning True. $\endgroup$ – Julián Aguirre Nov 28 '14 at 16:16
  • $\begingroup$ I see. I will edit the code. $\endgroup$ – Julián Aguirre Nov 28 '14 at 16:54

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