(i) By using your $r = 3 + 2\sqrt{2}$ of Example 8.10, determine $r^2$, $r^3$, and $r^4$. Check that these also yield solutions of $x^2 − 2y^2 = 1$.
(ii) Find approximations of $\sqrt{2}$ by using $r$, $r^2$, $r^3$, and $r^4$ and the discrepancy between $\sqrt{2}$ and your approximations correct to 3 significant figures.
(iii) Explain why the solutions to the equation $x^2 − 2y^2 = 1$ give rational approximations to $\sqrt{2}$.
(i) We start with
$$ r = 3 + 2\sqrt{2} $$
For $r^2$ we have
$$ r^2 = 3^2 + 2^2(2) + 3 \cdot 3 \cdot 2\sqrt{2} = 17 + 12\sqrt{2} $$
Validating $(x,y)=(17,12)$ is a solution
$$ 17^2 - 2(12^2) = 289 - 288 = 1$$
For $r^3$ we have
$$ r^3 = (3+2\sqrt{2})(17+12\sqrt{2}) = 51 + 24\cdot2 + 3\cdot12\sqrt{2} + 17\cdot2\sqrt{2} = 99 + 70\sqrt{2}$$
Validating $(x,y)=(99,70)$ is a solution
$$ 99^2 - 2(70^2) = 9801 - 9800 = 1 $$
For $r^4$ we have
$$r^4 = (r^2)^2 = (17+12\sqrt{2})^2 = 289 + 144\cdot2 + 2\cdot17\cdot12\sqrt{2} = 577 + 408\sqrt{2}$$
Validating $(x,y)=(577,408)$ is a solution
$$ 577^2 - 2(408^2) = 332929 - 332928 = 1$$
(ii) The following table shows approximations for $\sqrt{2}$ using $\frac{x}{y}$.
| r^n | x | y | x/y | x/y – sqrt(2) |
| r | 3 | 2 | 1.5 | 8.58E-02 |
| r^2 | 17 | 12 | 1.41666666666667 | 2.45E-03 |
| r^3 | 99 | 70 | 1.41428571428571 | 7.22E-05 |
| r^4 | 577 | 408 | 1.41421568627451 | 2.12E-06 |
The approximations improve as the power of $r$ grows.
(iii) Re-arranging $x^2-2y^2=1$ gives
$$ \frac{x}{y} = \sqrt{2+\frac{1}{y^2}} $$
And so the rational number $\frac{x}{y}$ better approximates $\sqrt{2}$ as $y$ grows larger. The solutions produced by $r^n$ as $n$ grows larger have larger values of $y$.