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Prove that if $2a^2 - b^2 = \pm1$ then $\frac ba\approx\sqrt2 $

(a,b) (1,1),(2,3),(5,7),(12,17),(29,41),(70,99)....

Steven Stadnicki
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Yu. K
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    "$\frac ba\approx\sqrt{2}$" is not a formal statement; if you want a proof then you'll need to define what you mean by 'approximately'. (You should also offer your thoughts on the problem; where it came from, what you've tried, etc.) – Steven Stadnicki Jan 05 '16 at 23:53
  • I suggest you read about continued fractions and Pell's equation. – Darth Geek Jan 05 '16 at 23:57
  • This is only true for sufficiently small values of $1$. :) – Deepak Jan 06 '16 at 00:49

2 Answers2

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Dividing both sides by $a^2$ and rearranging, we get $$2-\frac{\pm 1}{a^2}=\left(\frac{b}{a}\right)^2$$ As $a\to \infty$, the left hand side approaches $2$. Note that $\pm 1$ could be replaced by any integer of small absolute value. The only reason $\pm 1$ is important is if you want a precise statement about how good $\frac{b}{a}$ is as an approximant to $\sqrt{2}$.

vadim123
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$$2a^2-b^2=\pm1$$ implies $$2a^2=b^2\pm1$$ or $$\frac{b^2\pm1}{a^2}=\frac{b^2}{a^2}\pm\frac{1}{a^2}=2$$

So, we have $$\frac{b^2}{a^2}=2\pm \frac{1}{a^2}\approx 2$$

Therefore we have $\frac{b}{a}\approx \sqrt{2}$

Peter
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