5 is primitive root on modulo 47 then solve the $x^{20} \equiv21(mod47)$

Question

I have tried to look powers of 5 in modulo 47 but It seems not working.

lab bhattacharjee · Accepted Answer · 2020-06-12T12:16:57.117

1

As $5^3\equiv31\equiv-16,5^6\equiv(-16)^2\equiv21\pmod{47}$

Take discrete logarithm,

$20\cdot$ind$_5x\equiv$ind$_5{21}\pmod{46}\equiv6\equiv6-46$

ind$_5x\equiv-2\pmod{\dfrac{46}{(46,20)}}\implies x\equiv5^{23k-2}\pmod{47}$

Please use Solving a Linear Congruence to find $5^{-2}\pmod{47}$

edited Jun 12 '20 at 12:16

answered Jun 12 '20 at 11:31

lab bhattacharjee

1

Since $\gcd(20,46)=2$, there are two solutions. They are $15$ and $32$. – lhf Jun 12 '20 at 12:02
@lhf, You are correct, Rectified. – lab bhattacharjee Jun 12 '20 at 12:19

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