What is the sum of m and n?

Let $a = n-1,$ so the equation becomes $6m^2 = a^3-a.$ Let $y = 36m$ and $x = 6a$ ; then $y^2 = x^3 - 36x.$ Your favorite computer algebra package can find all the integral points on this elliptic curve; note that to give an integral solution to the original equation, you need $x$ to be a multiple of $6$ and $y$ a multiple of $36.$

MAGMA says that the integral points consist of: $(-6,0), (0,0), (6,0), (-3, \pm 9), (-2, \pm 8), (12,\pm 36), (18, \pm 72), (294, \pm 5040).$ Of these, the first five don't give meaningful solutions, the sixth gives $n=3, m=1,$ the seventh gives $n=4, m=2,$ and the eighth gives $n=50, m=140.$ This last solution is the only one that satisfies the problem conditions, so the answer is $\fbox{190}.$ If you believe that these are all the integral points, this shows that the solution is unique.

(There might be a nice descent argument to solve the equation as well, I don't know. It seems like a bit of a headache.)

$\binom n 3=m^2 \implies n(n-1)(n-2)=6.m^2$

I claim that there should be a one to one function between $(n,n-1,n-2)$ and three possible forms $(A^2,2B^2,3C^2)$ , where $A,B,C$ are positive integers. The argument is as below:

If there is a prime $p^\alpha$ ( $p\geq 5$ ) in one of $(n,n-1,n-2)$ , such that $\alpha$ is odd, then another member of $(n,n-1,n-2)$ should be divisible by $p^{\beta}$ ( $\beta$ is odd to), so that we make $m^2$ being a square. For the case $p=3$ , the same argument applies, except that power of $3$ should end up odd. But, we know that if $x$ is a multiple of $p$ , the next number, that is a multiple of $p$ , is $p\geq 5$ units away from $x$ , hence, integers $(n,n-1,n-2)$ , that are at most $3$ units away, can be of forms (not with exact order) $(2^{a}A^2,3^{b}B^2,C^2)$ or $(2^{a}A^2,3^{b}B^2,2^{c}C^2)$ or $(2^{a}3^{b}A^2,B^2,2^{b}C^2)$ , where $A,B,C$ are positive integers, that are coprimes with respect to $6$ . Note that $(2^{a}3^{b}A^2,B^2,C^2)$ is not a possible form, as $B^2$ and $C^2$ would become to squares, with maximum of $2$ units difference. The mentioned possible forms can all be simplified to $(2A^2,3B^2,C^2)$ , when we allow the even powers $2$ and $3$ to be absorbed by $A^2,B^2,C^2$ .

Knowing the general form $(A^2,2B^2,3C^2)$ , I wrote squares in one row and, in the row below, I wrote the same squares, multiplied by $2$ and in the next row, the squares, multiplied by $3$ . Then I looked for $3$ consecutive numbers, exactly from each and every row. There would be examples like $1,2,3$ or $2,3,4$ that should be ignored as they amount to an $n<7$ . The smallest triple, that would satisfy our conditions, is $(49,50,48)=(A^2,2B^2,3C^2)$ . I am not sure if there is gonna be another triple.

$n(n-1)(n-2)=50.49.48=117600=6.19600=6.(140)^2 \implies m+n=140+50=190$

$\sqrt{\binom{50}{3}} \Rightarrow 140$ There are no other solutions in positive integers for $\sqrt{\binom{n}{3}}$ and $n\geq 7$ .

This problem came from a Paul Ërdos proof. The book from which the problem was availed did not make the $n\geq 7$ restriction readily apparent and I missed that restriction.