From 'Proofs from the book', it stated that $ \tbinom n3=m^2$ has the unique solution n=50,m=140. But how do we prove this is so?
Expansion of the equation above yields $n(n-1)(n-2)=6m^2$ which can be factorised into $(n-1)^3-(n-1)=6m^2$ which is a diophantine equation. How do I prove that there are no other (positive) integer solutions for this equation?
Let $p\neq 2,3$ a prime divisor of $m$. Since $p$ can divide only one of $n$, $n-1$ or $n-2$, we conclude that, apart from square factors, the only possibilities for $(n,n-1,n-2)$ are: $$(6,1,1);(2,3,1);(2,1,3);(3,2,1);(1,2,3);(1,6,1);(3,1,2);(1,3,2);(1,1,6)$$ It`s simply to exclude (6,1,1) and (1,1,6), cause you can't have two consecutive square. In the case (2,3,1) you should have, in particular, the relation $a^2+1=3b^2$, and by standard theory on pell equation you know that there are no solution. In a similar way you exclude the cases (3,1,2) (1,2,3) and (1,6,1). Now consider the case (2,1,3). This is equivalent to the following system of pell equations: $$\begin{cases} b^2-3a^2=1\\ b^2-2c^2=-1 \end{cases}$$ Using again the theory on pell equation you can find the family of solution $(b_k,a_k)$ and $(b_k,c_k)$ for the two equations, and you search for a match of the $b_k$ of the two family. Since one of the sequence grows exponentially faster than the other, it suffices to verify just a few $k$. You find the only solution $a_k=4, b_k=7, c_k=5$, that corresponds to the triple $(50,49,48)$. In the case $(1,3,2)$ and $(3,2,1)$ you proceed in the same way and you find that there are no other solutions.