Is there a way to show $\frac{1646-736\sqrt{5}}{2641-1181\sqrt{5}}=\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$ without multiplying large numbers? I need to show that the following equality holds $$\frac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\frac{17+15\sqrt{5}} {7+15\sqrt{5}} $$ The only way I could prove it was via cross multiplication using a calculator.
Is there any simpler way to do this? Following prime factorization is given $$1646=2\times 823,736=2^5\times 23,2641=19\times 139$$ and $1181$ is prime. I strictly want to avoid multiplying large numbers. I don't know if that's possible here.
The above simplification is used to obtain a famous approximation to $\pi$ given by Ramanujan.
 A: Note that
\begin{eqnarray*}
\frac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =1 -5\frac{199-89\sqrt{5}}{2641-1181\sqrt{5}}
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{89\sqrt{5}-199}{2641-1181\sqrt{5}}=\frac{2}{7+15 \sqrt{5}}.
\end{eqnarray*}
A: There is another approach which is more in line with the answer linked in the question.
Let $\phi=(1+\sqrt {5})/2$ denote the golden ratio so that $\sqrt{5}=2\phi-1$ and $$\phi^2=\phi+1\tag{1}$$ The fraction in question can now be written as $$\frac{1191-736\phi}{1911-1181\phi}\tag{2} $$ Let us observe that $$ \frac{a-b\phi} {c-d\phi} =\frac{b\phi^2-a\phi}{d\phi^2-c\phi}=\frac{b-(a-b)\phi}{d-(c-d)\phi}\tag{3}$$ and apply this to $(2)$ repeatedly. We get the following sequence of fractions $$\frac{736-455\phi}{1181-730\phi}, \\ \frac{455-281\phi}{730-451\phi},\\ \frac{281-174\phi}{451-279\phi},\\ \frac{174-107\phi}{279-172\phi},\\ \frac{107-67\phi}{172-107\phi},\\ \frac{67-40\phi}{107-65\phi},\\ \frac{40-27\phi}{65-42\phi},\\ \frac{27-13\phi}{42-23\phi},\\ \frac{13-14\phi}{23-19\phi},\\ \frac{14-(-1)\phi}{19-4\phi},\\ \frac{-1-15\phi}{4-15\phi} $$ The last fraction equals $$\frac{15\phi+1}{15\phi-4}=\frac{30\phi+2}{30\phi-8}=\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$ The application of $(3)$ was stopped at a point when the numbers in the fraction were least in absolute value. Any further application would have led to larger numbers. This approach has advantage that we don't need to know beforehand the final simple answer.

If we have a fraction like $$\frac {a+b\phi} {c+d\phi} $$ then we can use the transformation $$\frac{a+b\phi} {c+d\phi} =\frac{a/\phi+b} {c/\phi+d} =\frac{a(\phi-1)+b}{c(\phi-1)+d}=\frac{b-a+a\phi}{d-c+c\phi}$$ and hope to get an expression with small numbers.
