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Question Number 193647 by Rupesh123 last updated on 17/Jun/23

	Answered by MM42 last updated on 17/Jun/23

	$b o t h n u m b e r s a, b c a n n o t b e o d d .$ $s o a = 2 o r b = 2$ $i f a = 2 & b = 3 \Rightarrow p = 2^{3} + 7 \times 3^{2} = 71 ✓$ $i f a = 3 & b = 2 \Rightarrow p = 3^{2} + 7 \times 2^{3} = 65 \times$ $i f a = 2 & b > 3; b i s p r i m e r n u m b e r$ $\Rightarrow b = 3 k + 1 \Rightarrow p = 2^{3 k + 1} + 7 \times {(3 k + 1)}^{2} \overset{3}{\equiv} 0 \times$ $\Rightarrow b = 3 k + 2 \Rightarrow p = 2^{3 k + 2} + 7 \times {(3 k + 2)}^{2} \overset{3}{\equiv} 0 \times$ $s i m i l a r y i n o t h e r c a s e s t h e r e s u l t$ $p i s n o t p r i m e n u m b e r$
	Answered by deleteduser1 last updated on 18/Jun/23

	$a a n d b c a n n o t b o t h b e o d d, i f s o, p w o u l d b e e v e n$ $\Rightarrow 2 = a^{b} + 7 b^{a} (a, b > 2) w h i c h i s i m p o s s i b l e$ $W h e n b = 2, p = a^{2} + 7 (2^{a})$ $a^{2} \equiv 1 (m o d 3), o t h e r w i s e a = 3 (g i v i n g p = 65 (c o n t r a d i c t i o n))$ $A l s o 7 (2^{a}) \equiv - 7 \equiv 2 (m o d 3) \Rightarrow p \equiv 0 (m o d 3)$ $\Rightarrow p = 3 = a^{2} + 7 (2^{a}) > 7 w h i c h i s i m p o s s i b l e$ $\Rightarrow b \neq 2 . . S o, w e c o n s i d e r a = 2 n e x t$ $W h e n a = 2 a n d b i s o d d p = 2^{b} + 7 b^{2}$ $b^{2} \equiv 1 (m o d 3) o t h e r w i s e b = 3 (g i v i n g p = 71)$ $S i n c e 71 i s a p r i m e, w e g e t (a, b) = (2, 3) p r o d u c e s$ $a p r i m e$ $C o n s i d e r i n g b^{2} \equiv 1 (m o d 3) g i v e s p = 3 (i m p o s s i b l e)$ $\Rightarrow T h e l a r g e s t s u c h p r i m e i s 71 .$
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