Let p be a prime number and d / p-1 . Then which of the following statements about the congruence? x ^d = 1( mod p) is / are correct : 1. It does not have a solution 2 atmost d incongurent solutions 3 exactly d incongruent solutions 4 aleast d incongruent solutions.
Hi Bhatia! What do you think? And why? (maybe try to work it out for a concrete example of p and d? Like p=11 and d=5?)
Thanks for your reply Going by your idea Let p= 5, then since d / p-1 then I guess we have d= 1, 2, 4...then solving x^ d = 1 (mod 5) for d=1, x=1 for d=2, x= 4, 6,9,11,14, .... for d=4, x= 2, 3,4,6, 7 ..... So I am thinking atleast d.... But not sure....I did not study number theory at graduation level.
You're working (mod 5) here, so I suggest that you express the x also in (mod 5). (thus x can be 0,1,2,3 or 4). When I do this, I get for d=1: x=1 for d=2: x=1,4 for d=4: x=1,2,3,4 Solutions like 9 are superfluous here, since 9=4 (mod 5). This is certainly correct, but the above example suggest exactly d! To prove this consider a number a coprime with p (for example take a=2 when p is odd). Then what can you say about [tex]a^{\frac{p-1}{d}}[/tex] By the way, do you know some group theory? It would make it a lot easier...
Thank you so much... I agree with this...I realized my mistake x should be strictly less than p. Taking p to be odd (as suggested) p= 5 , then d =1, 2, 3, 4. Does 2 ^ (p-1)/d imply the following we get: for d=1, 2^ 4 = 16 works for d=1 that is , 16= 1 (mod 5). for d=2, 2 ^ 2= 4 works for d=2 that is, 4^2 = 1( mod 5) for d= 4, 2 ^ 1 =2 works for d =4, that is , 2^ 4 = 1 (mod 5). Does this mean that we can generate other values of x ( above we get x= 16)....so x has to be at least d but can be more than d as well. Please help !