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If $x$ and $y$ are integers, then solve (using elementary methods)
$$3x^2-y^2=2$$

I tried the following
If $y$ is even, then $4|y^2$ and hence $2|y^2+2$ (and $4$ doesn't divide it), but $3x^2=y^2+2$ and since for R.H.S. to be even, $2|x^2 \implies 4|x^2$, and we get a contradiction. So $x$ and $y$ both are odd. If I substitute $x=2a+1$ and $y=2b+1$ for some integers $a$ and $b$, I get a more tedious equation and can't solve further. Please Help!
Thanks!

Bart Michels
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Henry
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  • This is the Generalized Pell Equation. For some solving techniques, you can see here, or try the Google. – rogerl Oct 28 '14 at 12:22
  • Does it help you make the proof if you know the solution?

    $$y = ±1/2 ((2-\sqrt3)^n+\sqrt3 (2-\sqrt3)^n+(2+\sqrt3)^n-\sqrt3 (2+\sqrt3)^n)$$ $$x = ±1/6 (3 (2-\sqrt3)^n+\sqrt3 (2-\sqrt3)^n+3 (2+\sqrt3)^n-\sqrt3 (2+\sqrt3)^n)$$

    – usethedeathstar Oct 28 '14 at 12:26
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    This isn't elementary, but here goes: $(1 - \sqrt{3})(1 + \sqrt{3}) = -2$. The fundamental unit in $\mathbb{Z}[\sqrt{3}]$ is $2 + \sqrt{3}$, therefore $(1 - \sqrt{3})(2 + \sqrt{3})^n$ will give us as many solutions as we want, including $x = 3, y = 5$; $x = 11, y = 19$; $x = 41, y = 71$; etc. – Robert Soupe Oct 28 '14 at 12:28
  • Hint: $2~(x^2-1)=y^2-x^2\iff2~(x-1)(x+1)=(y-x)(y+x)$. – Lucian Oct 28 '14 at 17:16
  • Okay, I'm confused. I tried $n = 2$ and that gives me -10.196152422706631880582339... Then I tried $n = 3$ and that got me -38.0525588832576502288019... How are you getting those nice integer answers from these irrational numbers? – Mr. Brooks Oct 29 '14 at 21:41
  • @Mr.Brooks An irrational number $a\sqrt3+b$ gives rise to a solution $(y,x)$ and vice versa. Note that $(a+\sqrt3b)(a-\sqrt3b)=a^2-3b^2$. I recommend you to study Pell (type) equations. – Bart Michels Oct 31 '14 at 20:54
  • @barto Oh, I see, thanks. – Mr. Brooks Nov 04 '14 at 15:51

1 Answers1

3

For the equation: $$3y^2-x^2=2$$

Knowing the first solution: $(y,x) - (3;5)$

The equation to find the rest:

$$y_2=x+2y$$

$$x_2=2x+3y$$

individ
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