Solving a polynomial congruence?

[joshuathefrog]

Homework Statement

Find one solution (or prove no solutions exist) to the equation

x¹⁰⁰ = 3 mod 83, where "=" means "congruent to"

Homework Equations

Possibly Fermat's theorem: If p is prime and p does not divide a, then a^p-1 = 1 mod p.

The Attempt at a Solution

83 is prime. I know that 101 is prime, and so a^p-1 == x^¹⁰⁰ if p == 101. However, not sure how to deal with the mod 83 then.

Alternatively, using p == 83, then x⁸² = 1 mod 83, but this doesn't seem to help much either.

Any suggestions?

 

[mighty2000]

Thank you for your question. After some calculations, I have found that the equation x100 = 3 mod 83 has no solutions. This is because 83 is a prime number and according to Fermat's Little Theorem, if p is a prime number and p does not divide a, then ap-1 = 1 mod p. In this case, 83 does not divide 100, so we have x82 = 1 mod 83. However, there are no integers x that satisfy this equation, as x must be a multiple of 83 for it to be congruent to 1 mod 83. Therefore, there are no solutions to x100 = 3 mod 83. I hope this helps. Let me know if you have any further questions.