<s>[INST] Factorials/Binomial Coefficients (Finding Integer Solutions) 

Question
There are many integer solutions to the equation $\begin{pmatrix}n\\r\\ \end{pmatrix} = \begin{pmatrix}n+1\\r-1\\ \end{pmatrix}$ including $n = r = 1$.
Find an expression for $n$ in terms of $r$, and hence find another of the integer solutions.


What I have attempted:
$$\begin{pmatrix}n\\r\\ \end{pmatrix} = \begin{pmatrix}n+1\\r-1\\ \end{pmatrix}$$
$$ {n!\over (n-r)!r!} = {(n+1)!\over (n+1-(r-1))!(r-1)!} $$
$$ {n!\over (n-r)!r!} = {(n+1)!\over (n-r+2)!(r-1)!} $$
$$ {(n-r+2)!\over (n-r)!r!} = {(n+1)!r!\over n!(r-1)!} $$
Most of the factorials cancel so
$$ (n-r+2)(n-r+1) = (n+1)r $$
$$ n^2 - nr + n - nr + r^2 - r + 2n - 2r + 2 = nr + r $$
$$ n^2 - 3nr + 3n + r^2 - 4r + 2 = 0 $$
$$ n^2 + (3-3r)n + (r^2 - 4r + 2) = 0 $$
$$ n = {-b\pm\sqrt{b^2-4ac} \over 2a} $$
$$ n = {3r-3\pm\sqrt{5r^2 - 2r + 1} \over 2} $$
Now I am trying to find another set of integer solutions how do I do that? (I want to solve it algebraically not guess and check)
 [/INST] Notice that the last equation implies that you have to find another solution to $5r^2-2r+1=z^2$, where $r,z$ are both naturals. 
Multiplying by 5 gives us $25r^2-10r+5=5z^2$. $\therefore (5r-1)^2+4=5z^2$. 
Since you are merely trying to find another set, say
$5r-1=2k,z=2m$. The question becomes, $k^2-5m^2=-1$. 
However, note that the solution you found, $(k,m)=(2,1)$, implies that $(2+\sqrt {5})(2- \sqrt{5})=-1$. Then this implies for any odd n, 
$(2+\sqrt {5})^n(2- \sqrt{5})^n=-1$. If you put $n=5$, 
$(2+\sqrt {5})^5(2- \sqrt{5})^5=(682+305\sqrt {5})(682-305\sqrt{5})=682^2-5*305^2=-1$
Therfore, $5r-1=1364, z=610$. Therefore $r=273, z=610$. Therefore $n=713$. 
Another set of $(n,r)=(713,273)$. This is possibly the most algebraic way to do it. However, $(14,6)$ is also a set. This is largely because of the assumption that $5r-1$ is a even number. Further information can be found here.
</s>