p premier p≥3 find p : 2^[(p-1)/2] ≡ 1[p]
- Thread starter RMOURAD
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What do you mean by "premier"?
Is there any particular meaning to the square brackets, such as the "floor" or "ceiling" function?
When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!
Eliz.
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Okay, so the question is:
Let [imath]p[/imath] be a prime number such that the following is true:
[imath]\qquad 2^{\normalsize{\frac{p-1}{2}}} \equiv 1\; (\textrm{mod }p)[/imath]
Thank you. Now, what are your thoughts? What have you tried? How far have you gotten?
Thank you!
blamocur
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My quick-and-dirty script found 11 primes under 100 satisfying the equation.
Why are you writing ................. p = 8k+ 1
When k = 3 or 4 , you would get p = 25 or 33 - which are not prime numbers
similarly for ................. p = 8k+ 7
When k = 1 or 4 , you would get p = 15 or 39 - which are not prime numbers