Show-that-if-3-prime-numbers-all-greater-than-3-form-an-arithmetic-progression-then-the-common-difference-of-the-progression-is-divisible-by-6-

Question Number 92673 by Ar Brandon last updated on 08/May/20

Show that if 3 prime numbers, all greater

than 3, form an arithmetic progression then the common

difference of the progression is divisible by 6 .

Commented by Rasheed.Sindhi last updated on 09/May/20

I n o t h e r w o r d s :

I f a & b a r e t w o p r i m e s s u c h t h a t

6 ∤ (b - a) t h e n 2 b - a i s c o m p o s i t e .

Commented by Ar Brandon last updated on 09/May/20

Interesting Sir. I'll like to get more details please.

Answered by Rasheed.Sindhi last updated on 10/May/20

L e t a, b \in P a r e f i r s t t w o t e r m s

∴ b - a i s c o m m o n d i f f e r e n c e

∴ T h e T h i r d i s b + (b - a) = 2 b - a

A c c o r d i n g t o t h e g i v e n i f (2 b - a) \in P

w i t h a, b \in P, t h e n 6 ∣ (b - a) .

I n o t h e r w o r d s i f 6 ∤ (b - a) t h e n

2 b - a i s n o t p r i m e i - e c o m p o s i t e .

P r o o f :

^{∙} A l l i n t e g e r s (w . r . t d i v i s i o n b y 6)

c a n b e c a t e g a r i z e d i n t o s i x t y p e s :

6 k, 6 k + 1, 6 k + 2, \dots, 6 k + 5 .

A m o n g t h e s e 6 k, 6 k + 2, 6 k + 3 &

6 k + 4 a r e o b v i o u s l y c o m p o s i t e .

T h i s m e a n s p r i m e s a r e o n l y o f

t w o t y p e s : 6 k + 1 & 6 k + 5

P o s s i b i l i t y f o r p r i m e p a i r (a, b)

a r e (6 k + 1, 6 k + 1), (6 k + 1, 6 k + 5)

(6 k + 5, 6 k + 1) & (6 k + 5, 6 k + 5)

a n d

b - a w i l l r e s p e c t i v e l y b e f a l l i n

t y p e s : 6 k, 6 k + 2, 6 k + 4 & 6 k .

T h a t m e a n s t h e p a i r s f o r w h i c h

a, b \in P a n d 6 ∤ (b - a)

(6 k + 1, 6 k + 5) & (6 k + 5, 6 k + 1)

A N D

{w e}^{'} v e t o p r o v e f o r t h e s e p a i r s

t h a t 2 b - a (t h e t h i r d t e r m) i s

n o t p r i m e i - e c o m p o s i t e . s e e

t h e t a b l e

| \begin{matrix} a & b & b - a & 2 b - a \\ 6 k + 1 & 6 k + 1 & \underset{6 ∣ (b - a)}{6 k} & 6 k + 1 \\ 6 k + 1 & 6 k + 5 & \underset{6 ∤ (b - a)}{6 k + 4} & \underset{N o n P r i m e}{6 k + 3^{}} \\ 6 k + 5 & 6 k + 1 & \underset{6 ∤ (b - a)}{6 k + 2} & \underset{N o n P r i m e}{6 k + 3^{}} \\ 6 k + 5 & 6 k + 5 & \underset{6 ∣ (b - a)}{6 k} & 6 k + 5 \end{matrix} |

^{} 6 k + 3 i s n o n p r i m e f o r k > 0

t h a t i s f o r a, b > 3

R e d r o w s r e p r e s e n t t h a t i f b - a

(c o m m o n d i f f e r e n c e) {c a n}^{'} t

b e d i v i d e d b y 6, t h e t h i r d t e r m

{c a n}^{'} t b e p r i m e .

T h a t i s i f t h r e e p r i m e n u m b e r s (> 3)

a r e i n A P, t h e n 6 m u s t d i v i d e c o m m o n

d i f f e r e n c e .

Q . E . D

Commented by Ar Brandon last updated on 10/May/20

Thank you ��