Given that the point (a,b) lies on the curve $x^2 - 3y^2=1$ find positive integers P,Q,R and S t (Pa + Qb, Ra + Sb) lies on the curve

Question

I am trying to solve this problem. However, I noticed that there are four unknowns while there are only 3 equations that I can use. As a result, guessing might involve.

I searched such equations on the internet, It said that it is called Pell's equation, and it can be solved using continued fraction. May I know how to use the continued fraction to solve simple pell's equation like this. Thank you very much.

score 1 · Accepted Answer · answered Jan 09 '20 at 22:21

1

$(x,y)=(2,1)$ is a solution; i.e., $2^2-3\cdot1^2=(2+\sqrt3)(2-\sqrt3)=1.$

If $(a,b) $ is a solution, i.e., $a^2-3b^2=(a+\sqrt3b)(a-\sqrt3b)=1, $

then $(2+\sqrt3)(a+\sqrt3b)(2-\sqrt3)(a-\sqrt3b)=1$ too.

$(2+\sqrt3)(a+\sqrt3b)=(2a+3b)+(a+2b)\sqrt3.$

That is, $(X,Y)=(2a+3b,a+2b)$ is also a solution of $X^2-3Y^2=1$.

I hope you find this answer useful, even though it doesn't use continued fractions.

answered Jan 09 '20 at 22:21

J. W. Tanner

60,406

more generally, if $a^2-3b^2=1$ and $c^2-3d^2=1$, then $(ac+3bd)^2-3(bc+ad)^2=(a^2-3b^2)(c^2-3d^2)=1$ – J. W. Tanner Jan 10 '20 at 03:43
Thank you very much; this is every useful. Would you mind to illustrate how to use continued fraction to solve this problem? ? – Henry Cai Jan 10 '20 at 09:31
@HenryCai: thanks for the acceptance; see this answer – J. W. Tanner Jan 10 '20 at 14:52

Given that the point (a,b) lies on the curve $x^2 - 3y^2=1$ find positive integers P,Q,R and S t (Pa + Qb, Ra + Sb) lies on the curve

1 Answers1