Congruences, Fermats Little Theorm

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SUMMARY

This discussion focuses on solving congruences using Fermat's Little Theorem. The first problem involves finding 'a' such that \(0 < a < 73\) and \(a \equiv 9^{794} \mod 73\). The solution utilizes the theorem to simplify the exponent, concluding that \(9^{794} \equiv 9^2 \mod 73\). The second problem, \(x^{86} \equiv 6 \mod 29\), is approached by recognizing that \(x^{28} \equiv 1 \mod 29\), leading to the simplification \(x^{86} \equiv x^2 \equiv 6 \mod 29\).

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Nebula
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Can someone explain how to solve these types of things, my book gives one example and its useless:

find a such that 0 < a < 73
a=9^794(mod 73) (= means congruent equals)

solve
x^86=6(mod 29)

Thanks in advance :smile:
 
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In the equation X^86 ==6 Mod 29, we note that for X not 0, X^28==1, by Fermat's Little Theorem. Thus X^86 ==X^28*X^28*X^28*X^2 ==X^2==6 Mod 29. That should be a start on the problem.
 
For the first problem :

You could start from the ground up :

9^2 = 81== 8 (mod 73) => 9^4 == 64 == -9 => 9^8==81==8...(lucky!) and so on, but you don't have to, because 73 is prime and does not divide 9. So you can happily use the Little Theorem.

ie: 9^72 == 1 (mod 73) => (9^72)^n == 1^n = 1 (mod 73), for any n. With n=11, you get 9^792==1 (mod 73).

So 9^794 == 9^2 (mod 73)
 
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