p premier p≥3 find p : 2^[(p-1)/2] ≡ 1[p]

[RMOURAD]

p premier p≥3
find p :

2^[(p-1)/2] ≡ 1[p]

 

[stapel]

p premier p≥3
find p :

2^[(p-1)/2] ≡ 1[p]

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.

 

[RMOURAD]

 

[stapel]

View attachment 35952

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!

 

[RMOURAD]

p = 8k +1 ، p = 8k +7
But I didn't get to the method

 

[blamocur]

p premier p≥3
find p :

2^[(p-1)/2] ≡ 1[p]

My quick-and-dirty script found 11 primes under 100 satisfying the equation.

 

[khansaheb]

p = 8k +1 ، p = 8k +7
But I didn't get to the method

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