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Variables a and b are integers: $b ≡ a \pmod {91}$ and $\gcd(a, 91) = 1$.

  • (a) Choose a positive number $k > 1$ so that $b^k ≡ a \pmod {91}$.

  • (b) What is $a \pmod {91}$ if $b = 53$?

Here is how I tried to do it:

  • a)

    $b^k ≡ a \mod 91$

    $b^{k-1}*b ≡ a*1 \pmod {91}$

    $b^{k-1} ≡ 1\pmod {91}$

    Now $\Phi (91) = 72$ gives $b^{72} = 1 \pmod {91}$ so k = 73.

  • b)

    $53^{73} = 53^{72} * 53 ≡ 53\ \pmod {91}$

I am wondering if I got it right and if I made any algebraic blunders.

Thank you

WimC
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noname197
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  • The answer to $a$ is correct. I have no idea what $b$ is asking..it appears to just be asking for $b\pmod {91}$, in which case your answer is correct though it has nothing to do with $73$ (but I suspect the problem was poorly transcribed). – lulu Jan 28 '21 at 13:28
  • Note: $73$ is not the smallest value you might have taken. $13$ works (exercise). But the problem statement doesn't appear to call for the minimal $k$. – lulu Jan 28 '21 at 13:30
  • I remember my professor published a paper for covering the general case and made an algorithm to solve this using computer –  Jan 28 '21 at 14:01
  • Part (b) makes no sense as posed. Please double check that you have not misquoted it. – Bill Dubuque Jan 28 '21 at 14:10

1 Answers1

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Hint:

Actually we need to use Carmichael Function

$\lambda(91)=[6,12]=12$ which should be the highest exponent for any integer $a,(a,91)=1$

Now $53\equiv-3\pmod7\implies53^3\equiv(-3)^3\equiv1\pmod7$

Again, $53\equiv1\pmod{13}$

$\implies53^{[1,3]}\equiv1\pmod{[13,7]}$

lab bhattacharjee
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