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Euler's theorem get the inverse of 2 in Z26...~ i tried to solve it but the result is not correct: a=2, φ (26) = 12, m=26 gcd(2,26)=21 multiplicative inverse of 2 = 1/2 2 * 1/2 = 1

gcd(2,26)=2

a^-1= a^11mod26=2^11mod26= 2046 mod 26= 20

a=2 a^-1=20 aa^-1= 220=40

when doing the verification, 40 mod 26 = 14 not 1...

or for this group the Euler theorem is not applicable ?

J. W. Tanner
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Eva
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    Welcome to Mathematics Stack Exchange. This would be easier to read if you used Mathjax. Euler's theorem assumes the base and modulus are relatively prime, but $\gcd(2,26)=2\ne1$ – J. W. Tanner Nov 26 '20 at 17:21
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    How did you get $\gcd(2,26)=21$ ? That's very wrong! – PM 2Ring Nov 26 '20 at 17:46
  • By the basic theorems in the linked dupes: $,a,$ is invertible $!\bmod n\iff \gcd(a,n)= 1.,$ If so, then Euler's theorem applies and can be used to compute the inverse. – Bill Dubuque Nov 27 '20 at 02:07

3 Answers3

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Welcome to MSE!

A nonzero element $a\in\Bbb Z_n$, $n\geq 2$, is invertible if and only if $gcd(a,n)=1$; otherwise $a$ is a zero divisor.

In your case $gcd(2,26)=2$ and so $2$ is not invertible.

Wuestenfux
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for this group the Euler theorem is not applicable ?

Euler's theorem assumes the base and modulus are relatively prime, but that is not the case here:

$\gcd(2,26)=2\ne1$.

J. W. Tanner
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$2$ cannot have an inverse mod $26$ because $13 \cdot 2 \equiv 0$.

Indeed, if $2a \equiv 1$, then $ 0 = 13 \cdot 2 a \equiv 13$, but $13 \not\equiv 0$.

lhf
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