Scientia Magna Vol. 3 (2007), No. 4, 104-104
A note on primes of the form a2 + 1 Juan L´opez Gonz´alez 9, December 2007 Madrid, Spain Abstract In this note I prove using an algebraic identity and Wilson’s Theorem that if a2 + 1 is an odd prime, thus this prime must has the form 4k2 + 1, then 5 - 2k − 3. Keywords Pseudo Smarandache function, mean value, asymptotic formula.
If n = a2 + 1 is prime and n 6= 2, then n is odd, thus a2 is even and n must has the form 4k + 1, where k ≥ 1 is an integer. The integers 4k 2 + 1 can be written as 2
4k 2 + 1 = (2k − 3)2 + 3(4k − 3) + 1.
(1)
If 2k − 3 = −1 then k = 1, and 5 - −1. If 2k − 3 = 1 then k = 2, 17 is a prime with (2 · 2 − 3, 5) = 1. If 2k − 3 > 1 then 4k 2 + 1 ≡ 0 + 3(2k) + 1( mod 2k − 3) ≡ 3(2k − 3) + 9 + 1( mod 2k − 3) ≡ 10 ( mod 2k − 3). By Wilson’s Theorem (4k 2 )! ≡ −1( mod 4k 2 + 1).
(2)
Thus exists an integer c such that (4k 2 )! + 1 = c · (4k 2 + 1), since 4k 2 > 2k − 3 for all k, then 2k − 3|(4k 2 )!, thus 0 + 1 ≡ c · 10( mod 2k − 3). (3) Then there are integers s and t, such that 10s + (2k − 3)t = 1,
(4)
thus (5, 2k − 3) = 1, by contradiction if 5|2k − 3, then 0 + 0 ≡ 1(mod 5). Thus I’ve proved the following Proposition. If a2 + 1 is an odd prime different of 5, then (a − 3, 5) = 1.
References [1] Richard K. Guy, Unsolved Problems in Number Theory, Springer, Second Edition, I(1994).