Show that there are infinitely many integers such that $ \binom{m}{n-1} = \binom{m-1}{n} $ This question comes from the 1st Brazilian's IMO TST of 2004. I have found no solutions of it online, though I have developed one.
After getting to $ mn = (m-n)(m-n+1) $, my solution relies on the claim that $ m_z = F(2z) \cdot F(2z+1) $ and $ n_z = F(2z-1) \cdot F(2z) $ (where $ F(z) $ is the Fibonacci function), which can be easily proved by finite induction.
Another attempt I had finds that $ n = \dfrac{3m+1 \pm \sqrt{5m^2+2m+1}}{2} $, but I couldn't prove that there are infinitely many integers $ m $ such that $ \sqrt{5m^2+2m+1} $ is an integer $ \equiv m+1 \pmod2 $.
Noticing $ 5m^2+2m+1 = (2m)^2 + (m+1)^2 $, I've tried setting a primitive Pythagorean triplet $ x^2-y^2=m+1, \; 2xy=2m $, but couldn't prove $ x^2-xy-y^2=1 $ has infinitely many integer solutions. I know that this is a hyperbola and that Pell can help me a lot here, but I've tried to restrain myself to using what I've learned in high school.
Can someone conclude my second approach or provide me a different~easier one?

Here’s an explicit approach to arriving at ali’s solution.
Apply the quadratic formula to $x^2-xy-y^2-1=0$, looking for non-negative solutions:
$$y=\frac{-x+\sqrt{x^2+4(x^2-1)}}2=\frac{-x+\sqrt{5x^2+4}}2\;.$$
Clearly this is an integer when $y\in\{0,1,3\}$, and a little more computation turns up the next integer value at $y=8$. Now look at the integer solutions discovered so far: $\langle 1,0\rangle,\langle 2,1\rangle,\langle 5,3\rangle$, and $\langle 13,8\rangle$. Those numbers are very familiar: turn the pairs around, and you have $0,1,1,2,3,5,8,13$, i.e., the Fibonacci numbers $F_0$ through $F_7$. If we denote successive solutions by $\langle x_n,y_n\rangle$, this immediately suggests that $y_{n+1}=x_n+y_n$ and $x_{n+1}=y_{n+1}+x_n=2x_n+y_n$. Once you guess those recurrences, it’s a trivial computation to verify that they do indeed yield solutions.
