The-number-of-pairs-of-natural-numbers-x-y-satisfy-x-2-y-gcd-x-y-2-2023-

Question Number 189531 by cortano12 last updated on 18/Mar/23

The number of pairs of natural

numbers (x, y) satisfy

x^{2} y + \gcd (x, y^{2}) = 2023

Answered by aleks041103 last updated on 19/Mar/23

l e t d = g c d (x, y)

\Rightarrow d ∣ x, y a n d d ∣ g c d (x, y^{2})

\Rightarrow d ∣ 2023 = 7 \times 17^{2}

x = a d, y = b d, g c d (a, b) = 1

\Rightarrow d^{3} a^{2} b + d g c d (a, b^{2} d) = 2023

\Rightarrow d^{3} < d^{3} a^{2} b < 2023 \Rightarrow d < \sqrt[3]{2023} \approx 12.65

\Rightarrow d ⩽ 12 a n d d ∣ 2023 = 7 . 17^{2}

\Rightarrow d = 1; 7

1 . d = 1

\Rightarrow g c d (x, y^{2}) = 1

\Rightarrow a^{2} b + 1 = 2023

\Rightarrow a^{2} b = 2022 = 2.3.337 \Rightarrow n o s o l t n .

2 . d = 7

\Rightarrow 49 a^{2} b + g c d (a, 7 b^{2}) = 17^{2} = 289

g c d (a, b) = 1 \Rightarrow g c d (a, 7 b^{2}) = g c d (a, 7)

\Rightarrow 49 a^{2} b + g c d (a, 7) = 289 = 17^{2}

g c d (a, 7) = 1; 7

i f g c d (a, 7) = 7 \Rightarrow 7 (7 a^{2} b + 1) = 17^{2} \Rightarrow n o s o l u t i o n

\Rightarrow g c d (a, 7) = 1

\Rightarrow 49 a^{2} b = 289 - 1 = 288

b u t 7 ∤ 288 \Rightarrow n o s o l u t i o n

\Rightarrow t h e r e i s n o s o l u t i o n \dots