Math Question and Answers


Question and Answers Forum

All Questions Topic List
Number Theory Questions
Previous in All Question Next in All Question
Previous in Number Theory Next in Number Theory
Question Number 182374 by rs4089 last updated on 08/Dec/22

Commented by Rasheed.Sindhi last updated on 08/Dec/22

$a = 10, b = 13, n = 3197$ $OR$ $a = 13, b = 10, n = 3197$
	Answered by Rasheed.Sindhi last updated on 09/Dec/22
	$n=a^3 +b^3 =(a+b)(a^2 −ab+b^2 ) ∵ a+b≤a^2 −ab+b^2 [see Q#182379] and 23 is the smallest prime factor of a^3 +b^3 ∴ a+b=23 with a^2 −ab+b^2 has no prime factor less than or equal to 23 Note: By {a_0 ,b_0 } we mean (a_0 ,b_0 ) or (b_0 ,a_0 ). Also we knowt If (a_0 ,b_(0 ) ) is solution so the (b_0 ,a_0 ). Case:{a,b}={1,22} a^2 −ab+b^2 =463=463 ✓ Case:{a,b}={2,21} a^2 −ab+b^2 =403=13_(<23) ×31 × Case:{a,b}={3,20} a^2 −ab+b^2 =349=349 ✓ Case:{a,b}={4,19} a^2 −ab+b^2 =301= 7_(<23) ×43 × Case:{a,b}={5,18} a^2 −ab+b^2 =259=7_(<23) ×37 Case:{a,b}={6,17} a^2 −ab+b^2 =223=223 ✓ Case:{a,b}={7,16} a^2 −ab+b^2 =193=193 ✓ Case:{a,b}={8,15} a^2 −ab+b^2 =169=13_(<23) ×13 Case:{a,b}={9,14} a^2 −ab+b^2 =151=151 ✓ Case:{a,b}={10,13}_(−) a^2 −ab+b^2 =139=139_(−) ✓ Case:{a,b}={11,12} a^2 −ab+b^2 =133=7_(<23) ×19 In underlined case a^2 −ab+b^2 has no prime factor≤23 and also 139 the least number. Hence 23×139=3197 is least value of n and {a,b}={10,13} n=a^3 +b^3 =(a+b)(a^2 −ab+b^2 )=23∙139=3197$
	$n = a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $You can't use 'macro parameter character #' in math mode$ $a n d 23 i s t h e s m a l l e s t p r i m e f a c t o r$ $o f a^{3} + b^{3}$ $∴ a + b = 23 w i t h a^{2} - a b + b^{2} h a s n o$ $p r i m e f a c t o r l e s s t h a n o r e q u a l t o$ $23$ $N o t e : B y {a_{0}, b_{0}} w e m e a n (a_{0}, b_{0}) o r (b_{0}, a_{0}) .$ $A l s o w e k n o w t I f (a_{0}, b_{0}) i s s o l u t i o n$ $s o t h e (b_{0}, a_{0}) .$ $Case : {a, b} = {1, 22}$ $a^{2} - a b + b^{2} = 463 = 463 ✓$ $Case : {a, b} = {2, 21}$ $a^{2} - a b + b^{2} = 403 = \underset{< 23}{13} \times 31 \times$ $Case : {a, b} = {3, 20}$ $a^{2} - a b + b^{2} = 349 = 349 ✓$ $Case : {a, b} = {4, 19}$ $a^{2} - a b + b^{2} = 301 = \underset{< 23}{7} \times 43 \times$ $Case : {a, b} = {5, 18}$ $a^{2} - a b + b^{2} = 259 = \underset{< 23}{7} \times 37$ $Case : {a, b} = {6, 17}$ $a^{2} - a b + b^{2} = 223 = 223 ✓$ $Case : {a, b} = {7, 16}$ $a^{2} - a b + b^{2} = 193 = 193 ✓$ $Case : {a, b} = {8, 15}$ $a^{2} - a b + b^{2} = 169 = \underset{< 23}{13} \times 13$ $Case : {a, b} = {9, 14}$ $a^{2} - a b + b^{2} = 151 = 151 ✓$ $\underline{Case : {a, b} = {10, 13}}$ $\underline{a^{2} - a b + b^{2} = 139 = 139} ✓$ $Case : {a, b} = {11, 12}$ $a^{2} - a b + b^{2} = 133 = \underset{< 23}{7} \times 19$ $In underlined case a^{2} - a b + b^{2} h a s$ $n o p r i m e f a c t o r ⩽ 23 a n d a l s o 139$ $t h e l e a s t n u m b e r .$ $H e n c e 23 \times 139 = 3197 i s l e a s t v a l u e$ $o f n a n d {a, b} = {10, 13}$ $n = a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) = 23 \cdot 139 = 3197$
	Answered by Rasheed.Sindhi last updated on 11/Dec/22

	$n = a^{3} + b^{3}; a, b \in N$ $S m a l l e s t p r i m e f a c t o r o f n = 23$ $\underline{a, b, n = ? i f n i s t h e s m l l e s t .}$ $L e t a^{3} + b^{3} = 23 p$ $w h e r e p (> 23) \in P a n d t h e p i s t h e l e a s t$ $(a + b) (a^{2} - a b + b^{2}) = 23 p$ $a + b = 23 \land a^{2} - a b + b^{2} = p$ $I f (a_{0}, b_{0}) s a t i s f i e s t h e n (b_{0}, a_{0}) a l s o d o e s .$ $b = 23 - a \land a^{2} - a (23 - a) + {(23 - a)}^{2} = p > 23$ $a^{2} - 23 a + a^{2} + a^{2} - 46 a + 529 > 23$ $3 a^{2} - 69 a + 529 > 23 \land 3 a^{2} - 69 a + 529 \in P$ $L e t f (a) = 3 a^{2} - 69 a + 529$ $f (1) = 463 \in P ✓$ $f (2) = 403 \notin P$ $f (3) = 349 \in P ✓$ $f (4) = 301 \notin P$ $f (5) = 259 \notin P$ $f (6) = 223 \in P ✓$ $f (7) = 193 \in P ✓$ $f (8) = 169 \notin P$ $f (9) = 151 \in P ✓$ $f (10) = 139 \in P ✓ t h e l e a s t$ $f (11) = 133 \notin P$ $a = 10 \Rightarrow b = 23 - 10 = 13$ $n = a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) = 23.139 = 3197$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum