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?

To prove that $x^2-xy-y^2=1$ has infinite integer solutions, you can see $ (1,0) $ is a solution; and if $ (a,b) $ is a solution, $ (2a+b,a+b) $ also is. You can find this relation by observing small cases.
