If you take the continued fraction out of $\sqrt7$, you get [2;1, 1, 1, 4...] which yields: $$2+\frac1{1+\frac1{\frac11+1}}=\frac8{3}$$ and indeed, $$8^2-7⋅3^2=1$$ However, if you take the continued fraction out of $\sqrt13$ you get [3;1, 1, 1, 1, 6...] which yields: $$3+\frac1{1+\frac1{1+\frac1{\frac11+1}}}=\frac{18}5$$ but that gets $$18^2-13⋅5^2=\textbf{-1}$$not $\textbf{1}$ but $\textbf{-1}$! Why? Where is my mistake?
2025-01-13 02:17:51.1736734671
Problem in a Pell equation
170 Views Asked by Ran https://math.techqa.club/user/ran/detail At
2
If you take a prime $p \equiv 1 \pmod 4$ you are guaranteed to get $-1$ first, but get back to $1$ if you repeat the periodic part of the CF.
Note $$18^2 + 13 \cdot 5^2 = 649,$$ $$ 2 \cdot 18 \cdot 5 = 180, $$ $$ 649^2 - 13 \cdot 180^2 = 1. $$
The continued fraction for $\sqrt {13}$ in the display I like. For a convergent $p/q,$ the number directly below is $p^2 - 13 q^2.$ This includes the fake convergent $1/0,$ which needs to be there to start the process.
$$ \small \begin{array}{cccccccccccccccccccccccccccccc} & & 3 & & 1 & & 1 & & 1 & & 1 & & 6 & & 1 & & 1 & & 1 & & 1 & & 6 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{3}{1} & & \frac{4}{1} & & \frac{7}{2} & & \frac{11}{3} & & \frac{18}{5} & & \frac{119}{33} & & \frac{137}{38} & & \frac{256}{71} & & \frac{393}{109} & & \frac{649}{180} & & \frac{4287}{1189} \\ \\ & 1 & & -4 & & 3 & & -3 & & 4 & & -1 & & 4 & & -3 & & 3 & & -4 & & 1 & & -4 \end{array} $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$ $$ 2^2 - 5 \cdot 1^2 = -1,$$ $$2^2 + 5 \cdot 1^2 = 9,$$ $$ 2 \cdot 2 \cdot 1 = 4, $$ $$ 9^2 - 5 \cdot 4^2 = 1. $$ $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$ $$ 70^2 - 29 \cdot 13^2 = -1,$$ $$70^2 + 29 \cdot 13^2 = 9801,$$ $$ 2 \cdot 70 \cdot 13 = 1820, $$ $$ 9801^2 - 29 \cdot 1820^2 = 1. $$
I put a complete picture for $\sqrt {29}$ at Solving Diophantine Equation - odd Periods