If-p-is-a-prime-number-greater-than-5-the-prove-that-p-mod-6-1-or-p-mod-6-5-i-e-All-prime-numbers-greater-than-5-leave-a-remainder-of-1-or-5-when-divided-by-6-

Question Number 4942 by prakash jain last updated on 24/Mar/16

If p is a prime number greater than 5, the prove

that p \mod 6 \equiv 1 or p \mod 6 \equiv 5 .

i . e . All prime numbers greater than 5 leave a

remainder of 1 or 5 when divided by 6 .

Commented by Yozzii last updated on 25/Mar/16

D e f i n e t h e f u n c t i o n f : P \to N s u c h t h a t

f (p) = p (m o d 6) w h e r e 0 ⩽ f (p) ⩽ 6 a n d p > 5 .

S i n c e p \in P, ∄ n \in N s u c h t h a t p = 6 n + 2 = 2 (3 n + 1) (> 5),

p = 6 n + 3 = 3 (2 n + 1) (> 5), p = 6 n + 4 = 2 (3 n + 2) (> 5) o r p = 6 n = 2 \times 3 n (> 5) . T h e r e f o r e,

w e k n o w t h a t f (p) \neq 0, 2, 3, 4 a n d s o,

p o s s i b l y f (p) = 1 o r f (p) = 5 s i n c e f o r

p = 6 n + 1 o r p = 6 n + 5, n o i n t e g e r f a c t o r s

a r i s e f r o m t h e s e f o r m u l a e o f p .

I f n i s e v e n, f (6 n + 1) = 1 o r f (6 n + 5) = 5 .

I f n i s o d d \Rightarrow p = 6 (2 k + 1) + 1 = 12 n + 7

\Rightarrow f (12 k + 7) = 1 o r i f p = 6 (2 k + 1) + 5 = 12 k + 11

\Rightarrow f (12 k + 11) = 5 .

Commented by 3 last updated on 15/May/16

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