# Convergents as solutions for Pell's equation

We know that convergents of $\sqrt 2$ offer solutions for Pell's equation of the format ${x^2} - 2{y^2} = 1$. I wonder how rational approximations of an irrational number become integer solutions for the above equation?

Also why only some convergents are integer solutions and why some or not? For example, $\boxed{\frac{3}{2}},\dfrac{7}{5}$, $\boxed{\frac{17}{12}},\dfrac{41}{29}$, . . . only some of the above are solutions and some are not. Is there any rule that governs which convergents become solutions? Can any one throw some light?

• Oct 1, 2016 at 15:13
• the others give $x^2 - 2 y^2 = -1.$ Oct 1, 2016 at 17:11

Here’s a statement of results, completely without proof:

If you calculate the continued fraction for $\sqrt n$, where $m^2<n<(m+1)^2$, you’ll find that it takes the form $m+\frac1{a_1+}\frac1{a_2+}\cdots\frac1{a_2+}\frac1{a_1+}\frac1{2m+}\cdots$, and the whole symmetric sequence repeats infinitely after that.

It happens too that all the $a_i$ are in the range $1\le a_i\le m$, and in fact $a_i=m$ happens (if it does at all) only in the very middle of the sequence of $a_i$’s. At any rate, your solutions of the Pell Equation $X^2-nY^2=\pm1$ come only from the convergents that you get by cutting off your continued fraction just before the appearance of the $2m$.

As an example, the expansion of $\sqrt{14}$ is $3+\frac1{1+}\frac1{2+}\frac1{1+}\frac1{6+}\frac1{1+}\frac1{2+}\frac1{1+}\frac1{6+}\cdots$, and to get your solutions of Pell, you cut off just before the $\frac16$. The first one is $3+\frac1{1+}\frac1{2+}\frac11=\frac{15}4$, and surenough, $X=15$, $Y=4$ is a solution.

Want to get only the solutions $X^2-nY^2=+1$? Depends on the length of the repeating part of the expansion. If it’s even, all the valid convergents give solutions, while if it’s odd as with $\sqrt2=1+\frac1{2+}\cdots$, the first valid convergent gives $-1$, the second gives $+1$, the third valid convergent gives $-1$, and so on.

This is $\sqrt 2$ and $x^2 - 2 y^2.$ $$\small \begin{array}{cccccccccccccccccccccccccccccc} & & 1 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{1}{1} & & \frac{3}{2} & & \frac{7}{5} & & \frac{17}{12} & & \frac{41}{29} & & \frac{99}{70} & & \frac{239}{169} & & \frac{577}{408} & & \frac{1393}{985} & & \frac{3363}{2378} & & \frac{8119}{5741} \\ \\ & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & & -1 \end{array}$$ ......................................

For other positive, squarefree $n,$ the convergents have $x^2 - n y^2$ alternating signs, giving values with absolute value smaller than $2 \sqrt n$

This is $\sqrt 3$ and $x^2 - 3 y^2.$ $$\small \begin{array}{cccccccccccccccccccccccccccccc} & & 1 & & 1 & & 2 & & 1 & & 2 & & 1 & & 2 & & 1 & & 2 & & 1 & & 2 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{1}{1} & & \frac{2}{1} & & \frac{5}{3} & & \frac{7}{4} & & \frac{19}{11} & & \frac{26}{15} & & \frac{71}{41} & & \frac{97}{56} & & \frac{265}{153} & & \frac{362}{209} & & \frac{989}{571} \\ \\ & 1 & & -2 & & 1 & & -2 & & 1 & & -2 & & 1 & & -2 & & 1 & & -2 & & 1 & & -2 \end{array}$$

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This is $\sqrt {14}$ and $x^2 - 14 y^2.$ $$\small \begin{array}{cccccccccccccccccccccccccccccc} & & 3 & & 1 & & 2 & & 1 & & 6 & & 1 & & 2 & & 1 & & 6 & & 1 & & 2 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{3}{1} & & \frac{4}{1} & & \frac{11}{3} & & \frac{15}{4} & & \frac{101}{27} & & \frac{116}{31} & & \frac{333}{89} & & \frac{449}{120} & & \frac{3027}{809} & & \frac{3476}{929} & & \frac{9979}{2667} \\ \\ & 1 & & -5 & & 2 & & -5 & & 1 & & -5 & & 2 & & -5 & & 1 & & -5 & & 2 & & -5 \end{array}$$