<s>[INST] Problem in a Pell equation 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?
 [/INST] 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 
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