= Solution
The complete <Butcher tableau> in the PDF supplies
$$
s=\frac{\sqrt3}{6},\qquad
A=\begin{pmatrix}\frac14&\frac14-s\\\frac14+s&\frac14\end{pmatrix},\quad
b=\frac12\begin{pmatrix}1\\1\end{pmatrix},\quad
c=\begin{pmatrix}\frac12-s\\\frac12+s\end{pmatrix}.
$$
These are the two Gauss nodes and the corresponding <Gauss--Legendre Runge-Kutta method>. We can verify its order directly rather than infer it from its name. Write $e=(1,1)^T$, $C=\operatorname{diag}(c)$ and let powers of $c$ be componentwise. The <Butcher order conditions> through order four are
$$
\begin{aligned}
b^Te&=1,&b^Tc&=\frac12,\\
b^Tc^2&=\frac13,&b^TAc&=\frac16,\\
b^Tc^3&=\frac14,&b^TCAc&=\frac18,\\
b^TAc^2&=\frac1{12},&b^TA^2c&=\frac1{24}.
\end{aligned}
$$
For these coefficients $Ae=c$, $Ac=c^2/2$ and $c^2=c-e/6$. Also $b^Tc^j=1/(j+1)$ for $j=0,1,2,3$, as follows by averaging the two values $1/2\pm s$. These identities give all eight displayed <Butcher order conditions>; for example $b^TCAc=b^Tc^3/2=1/8$, $b^TAc^2=b^T(c^2/2-c/6)=1/12$, and $b^TA^2c=b^TAc^2/2=1/24$. They establish order at least four for smooth nonlinear <ordinary differential equations>.
To exclude order five, apply the method to the <Dahlquist test equation>. Solving its stage equations gives the <stability function>
$$
R(z)=1+zb^T(I-zA)^{-1}e
=\frac{1+z/2+z^2/12}{1-z/2+z^2/12}
=1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{144}+O(z^6).
$$
The exact solution has coefficient $1/120$ at degree five. Hence \b[the method has order exactly four]; its one-step defect is $O(h^5)$.
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