> *************
> Proth's Theorem (1878): Let n = k.(2^q)+1 with 2^q > k. If there is
> an integer $a$ such that a^((n-1)/2) = -1 (mod n), then $n$ is prime.
> *************
>
> Edouard Lucas claims he stated for the first time in 1876 the following
> (famous) theorem:
>
> *If a^x = 1 (mod n) for x = n-1 and for no other divisor of (n-1),
> then the number $n$ is prime.*
>
> of which Proth's is a quite obvious consequence.
To which Avinoam Mann remarked:
> While both theorems are not difficult to prove, I am not sure that Proth's
> follows from Lucas'.
My answer:
Proth's is said to be an obvious consequence of Lucas' for:
(n-1)/2 = k.(2^q)/2, and 2^q > k insure that (n-1)/2 is even (yes, it
does, except for n=3), then a^((n-1)/2) = -1 (mod n) fits perfectly with
Lucas' conditions.
May I ask Avinoam Mann for "not difficult" proofs of both theorems? (quote:
"both theorems are not difficult to prove")
Meanwhile, I shall read in the Comptes Rendus Acad. Sci. Paris, LXXXVII,
p. 374, 1878. (as Alfred Ross indicated it could be found) for Proth's own
proof.
Udai Venedem
venedem@wanadoo.fr
http://perso.wanadoo.fr/alta.mathematica/