
Test
Hi, you'll have to learn how to calculate square roots with continued fractions.
Since formatting continued fractions is a nightmare of parentheses, I'll refer you to Continued Fraction  from Wolfram MathWorld and ask you to read equation (3), which gives the basic form of the continued fractions we'll need and the shortcut form for this (4) so I don't scratch my eyeballs out trying to format things. Also see (11) for the finite version of this. Another bit of slightly confusing notation, [x] will mean the greatest integer less than x. This shouldn't be confused with our continued fraction notation, since that will always have more than one term.
Assume D is not a perfect square. To find the continued fraction expression of \sqrt{D}, we first set a_{0}=[\sqrt{D}]. This is a very crude approximation to \sqrt{D}. At this point we have \sqrt{D}=a_{0}+(\sqrt{D}a_{0})=a_{0}+\frac{1}{\frac{1}{\sqrt{D}a_{0}}}
We apply the same procedure to \frac{1}{\sqrt{D}a_{0}}=\frac{\sqrt{D}+a_{0}}{Da_{0}^{2}} and get a_{1}=[\frac{\sqrt{D}+a_{0}}{Da_{0}^{2}}]
Now we have \sqrt{D}=a_{0}+\frac{1}{a_{1}+(\frac{\sqrt{D}+a_{0 }}{Da_{0}^{2}}a_{1})}
Now a_{2}=[\frac{1}{\frac{\sqrt{D}+a_{0}}{Da_{0}^{2}}a_{1}}]. Continue to get the rest of the a's. Eventually you'll get something that repeats like \sqrt{D}=[a_{0}, a_{1}, \ldots, a_{k}, a_{0}+\sqrt{D}]
Heres where you stop. If k is odd find integers x and y where x/y=[a_{0}, a_{1}, \ldots, a_{k}]. These are your minimal solutions to Pell's equation. If k is even, you do something similar, I'm not positive exactly what, sorry. I'll hopefully come back tomorrow with an answer.
An example: D=14
a_{0}=[\sqrt{14}]=3 so \sqrt{14}=3+\frac{1}{\frac{1}{\sqrt{14}3}}
a_{1}=[\frac{1}{\sqrt{14}3}]=[\frac{\sqrt{14}+3}{5}]=1
So \sqrt{14}=3+\frac{1}{1+(\frac{\sqrt{14}+3}{5}1)}=3+\frac{1}{1+\frac{1}{\frac{1}{\frac{\sqrt{14} +3}{5}1}}}
a_{2}=[\frac{1}{\frac{\sqrt{14}+3}{5}1}]=[\frac{5}{\sqrt{14}2}]=[\frac{\sqrt{14}+2}{2}]=2
So
\sqrt{14}=3+\frac{1}{1+\frac{1}{2+(\frac{\sqrt{14} +2}{2}2)}}
ok I'm stopping here. Go a couple more steps and you'll get \sqrt{14}=[3,1,2,1,3+\sqrt{14}], so we've started to repeat.
Now find [3,1,2,1]=3+\frac{1}{1+\frac{1}{2+\frac{1}{1}}}=15/4, so x=15 and y=4 are the minimal solutions in this case.