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.


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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

	$L e t a, b \in P a r e f i r s t t w o t e r m s$ $∴ b - a i s c o m m o n d i f f e r e n c e$ $∴ T h e T h i r d i s b + (b - a) = 2 b - a$ $A c c o r d i n g t o t h e g i v e n i f (2 b - a) \in P$ $w i t h a, b \in P, t h e n 6 ∣ (b - a) .$ $I n o t h e r w o r d s i f 6 ∤ (b - a) t h e n$ $2 b - a i s n o t p r i m e i - e c o m p o s i t e .$ $P r o o f :$ $^{∙} A l l i n t e g e r s (w . r . t d i v i s i o n b y 6)$ $c a n b e c a t e g a r i z e d i n t o s i x t y p e s :$ $6 k, 6 k + 1, 6 k + 2, . . ., 6 k + 5 .$ $A m o n g t h e s e 6 k, 6 k + 2, 6 k + 3 &$ $6 k + 4 a r e o b v i o u s l y c o m p o s i t e .$ $T h i s m e a n s p r i m e s a r e o n l y o f$ $t w o t y p e s : 6 k + 1 & 6 k + 5$ $P o s s i b i l i t y f o r p r i m e p a i r (a, b)$ $a r e (6 k + 1, 6 k + 1), (6 k + 1, 6 k + 5)$ $(6 k + 5, 6 k + 1) & (6 k + 5, 6 k + 5)$ $a n d$ $b - a w i l l r e s p e c t i v e l y b e f a l l i n$ $t y p e s : 6 k, 6 k + 2, 6 k + 4 & 6 k .$ $T h a t m e a n s t h e p a i r s f o r w h i c h$ $a, b \in P a n d 6 ∤ (b - a)$ $(6 k + 1, 6 k + 5) & (6 k + 5, 6 k + 1)$ $A N D$ ${w e}^{'} v e t o p r o v e f o r t h e s e p a i r s$ $t h a t 2 b - a (t h e t h i r d t e r m) i s$ $n o t p r i m e i - e c o m p o s i t e . s e e$ $t h e t a b l e$ $\| \begin{matrix} a & b & b - a & 2 b - a \\ 6 k + 1 & 6 k + 1 & \underset{6 ∣ (b - a)}{6 k} & 6 k + 1 \\ 6 k + 1 & 6 k + 5 & \underset{6 ∤ (b - a)}{6 k + 4} & \underset{N o n P r i m e}{6 k + 3^{◼}} \\ 6 k + 5 & 6 k + 1 & \underset{6 ∤ (b - a)}{6 k + 2} & \underset{N o n P r i m e}{6 k + 3^{◼}} \\ 6 k + 5 & 6 k + 5 & \underset{6 ∣ (b - a)}{6 k} & 6 k + 5 \end{matrix} \|$ $^{◼} 6 k + 3 i s n o n p r i m e f o r k > 0$ $t h a t i s f o r a, b > 3$ $R e d r o w s r e p r e s e n t t h a t i f b - a$ $(c o m m o n d i f f e r e n c e) {c a n}^{'} t$ $b e d i v i d e d b y 6, t h e t h i r d t e r m$ ${c a n}^{'} t b e p r i m e .$ $T h a t i s i f t h r e e p r i m e n u m b e r s (> 3)$ $a r e i n A P, t h e n 6 m u s t d i v i d e c o m m o n$ $d i f f e r e n c e .$ $Q . E . D$
	Commented by Ar Brandon last updated on 10/May/20
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