Diophantine method

Question

I am having a problem in understanding the following problem:

Find $\sqrt{15}$ using Diophantine method.

I am aware of what Diophantine equations are, but totally stuck when asked to find $\sqrt{15}$, with $3$ decimal places.

What does it mean?

Answer 1

The Pell equation method ends up with this: given a pair of (non-negative) integers $x,y$ that satisfy $x^2 - 15 y^2 = 1,$ we get as many solutions as might be needed by applying $$ (x,y) \mapsto (4x+15y, x+4y). $$ The ratios get closer and closer to $\sqrt{15}$ since $$ \left( \frac{x}{y} \right)^2 - 15 = \frac{1}{y^2} $$ $$ \left( \frac{x}{y} - \sqrt{15} \right) \left( \frac{x}{y} + \sqrt{15} \right) = \frac{1}{y^2} $$ $$ \left( \frac{x}{y} - \sqrt{15} \right) = \frac{1}{y^2 \left( \frac{x}{y} + \sqrt{15} \right)} $$ So the error is $ \frac{1}{y^2 \left( \frac{x}{y} + \sqrt{15} \right)}, $ very close to $\frac{1}{ 2 \sqrt{15}y^2 }$ and less than $\frac{1}{7 y^2}$ We get $$ (4,1) , 4.0 $$ $$ (31, 8) , 3.875 $$ $$ (244,63), \; 3.87301 $$ $$ (1921,496) , \; 3.87298 $$ $$ (15124,3905), \; 3.872983355 $$ $$ (119071, 30744), \; 3.872983346 $$

If we switch to a negative target, the ratios are slightly smaller than $\sqrt{15}.$ Using $-6,$ the errors come out $ \frac{-6}{y^2 \left( \frac{x}{y} + \sqrt{15} \right)}, $ very close to $\frac{-6}{ 2 \sqrt{15}y^2 }$ and less than $\frac{6}{7 y^2}$ in absolute value... $$ (3,1) , 3.0 $$ $$ (27, 7) , 3.857 $$ $$ (213,55), \; 3.8727 $$ $$ (1677,433) , \; 3.872979 $$ $$ (13203,3409), \; 3.872983 $$ $$ ( 103947 , 26839 ), \; 3.872983345 $$