Question

Solve in positive integers $x^2-13y^2=\pm 1$

• A. $x=18,y=5$
• B. $x=5,y=18$
• C. $x=180,y=649$
• D. $x=649,y=180$

Answer 1

Answer: (A), (D)

The correct options are
A $x=18,y=5$
D $x=649,y=180$

$x^2+13y^2=\pm 1$

$x^2+13y^2=-1$

Let $y=1$ & $x=4$

$4^2-13.1^2=3$

Let $y=2$ & $x=7$

$7^2-13.2^2=-3$

We know that,

$(b^2-n{a}^2)(d^2-n{c}^2)={(bd+nac)}^2-n{(bc+ad)}^2$

Here $(a,b)=(1,4)$ & $(c,d)=(2,7),n=13$

$\left(3\right)(-3)={(28+13.\left(1\right)\left(2\right))}^2-13{(\left(4\right)\left(2\right)+\left(1\right)\left(7\right))}^2$

$-9={\left(54\right)}^2-13{\left(15\right)}^2$

$-1={\left(10\right)}^2-13{\left(5\right)}^2$

$\therefore x=18$ & $y=5$ is solution for $x^2-13y^2=-1$

$\therefore {18}^2-13{\left(5\right)}^2=-1$

$({18}^2-13{\left(5\right)}^2)({18}^2-13{\left(5\right)}^2)={(\left(18\right)\left(18\right)+13\left(5\right)\left(5\right))}^2-13{(18\left(5\right)+5\left(18\right))}^2$

$(-1)(-1)={\left(649\right)}^2-13{\left(180\right)}^2$

$\therefore {\left(649\right)}^2-13{\left(180\right)}^2=1$

$\therefore x=649,y=180$ is solution for $x^2-13y^2=1$