Let (√a)+ (√b)= (√(2023)) , Find values of a, b ∈ N


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Question Number 178596 by Acem last updated on 18/Oct/22

$L e t \sqrt{a} + \sqrt{b} = \sqrt{2023}, F i n d v a l u e s o f a, b \in N$
Commented by mr W last updated on 18/Oct/22

$a = 7 {(17 - n)}^{2}$ $b = 7 n^{2}$ $w i t h n = 1, 2, . . ., 16$
Commented by Acem last updated on 18/Oct/22

$I l i k e e l e g a n t s o l u t i o n s!$ $T m o r r o w i w i l l t r y t h i s q u s e t i o n$
Commented by mr W last updated on 19/Oct/22

$s i n c e 2023 = 7 \times 17^{2}, s o i n t h i s c a s e$ $t h e s o l u t i o n i s v e r y o b v i o u s .$ $\sqrt{a} + \sqrt{b} = 17 \sqrt{7}$ $a = 7 A^{2}, b = 7 B^{2}$ $A + B = 17$ $l e t B = n, t h e n A = 17 - n$ $\Rightarrow a = 7 {(17 - n)}^{2}, b = 7 n^{2}$
Commented by mr W last updated on 19/Oct/22

$a n d w e c a n a l s o s e e$ $\sqrt{a} + \sqrt{b} = \sqrt{2022} h a s n o s o l u t i o n, s i n c e$ $2022 = 2 \times 3 \times 337 w h i c h {d o e s n}^{'} t c o n t a i n$ $a p e r f e c t s q u a r e .$
	Answered by Rasheed.Sindhi last updated on 19/Oct/22
	${ (((√a) +(√b) =17(√7) ....(i))),(((√a) −(√b) =u(√7) (say)....(ii))) :}[(√a) ,(√b) ≤17(√7) ] (i)+(ii): (√a) =(((17+u)(√7))/2) a=((7(17+u)^2 )/4)≤; u∈O a=((7(17+(2n+1))^2 )/4); n∈Z a=((7(18+2n)^2 )/4); n∈Z determinant ((( a=7(9+n)^2 ≤17(√7) ; n∈Z))) (i)−(ii):(√b) =(((17 −(2n+1))(√7))/2) b=((7(17 −u)^2 )/4); u∈O b=((7(17 −(2n+1))^2 )/4); n∈Z determinant (((b=7(8 −n)^2 ≤17(√7) ; n∈Z))) Continue....$
	${\begin{cases} \sqrt{a} + \sqrt{b} = 17 \sqrt{7} . . . . (i) \\ \sqrt{a} - \sqrt{b} = u \sqrt{7} (s a y) . . . . (i i) \end{cases} [\sqrt{a}, \sqrt{b} ⩽ 17 \sqrt{7}]$ $(i) + (i i) : \sqrt{a} = \frac{(17 + u) \sqrt{7}}{2}$ $a = \frac{7 {(17 + u)}^{2}}{4} ⩽; u \in O$ $a = \frac{7 {(17 + (2 n + 1))}^{2}}{4}; n \in Z$ $a = \frac{7 {(18 + 2 n)}^{2}}{4}; n \in Z$ $\begin{matrix} a = 7 {(9 + n)}^{2} ⩽ 17 \sqrt{7}; n \in Z \end{matrix}$ $(i) - (i i) : \sqrt{b} = \frac{(17 - (2 n + 1)) \sqrt{7}}{2}$ $b = \frac{7 {(17 - u)}^{2}}{4}; u \in O$ $b = \frac{7 {(17 - (2 n + 1))}^{2}}{4}; n \in Z$ $\begin{matrix} b = 7 {(8 - n)}^{2} ⩽ 17 \sqrt{7}; n \in Z \end{matrix}$ $C o n t i n u e . . . .$
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