Let-a-b-2023-Find-values-of-a-b-N-

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, \dots, 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

{\begin{cases} \sqrt{a} + \sqrt{b} = 17 \sqrt{7} \dots . (i) \\ \sqrt{a} - \sqrt{b} = u \sqrt{7} (s a y) \dots . (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 \dots .