Solutions to $\binom{n}{5} = 2 \binom{m}{5}$

Many Diophantine equations are solved using modern algebraic geometry. For an informal survey how this works, see

M. Stoll, How to solve a Diophantine equation, arXiv.

The most prominent example is Fermat's equation. But there are also interesting binomial equations. It has been shown very recently that $\binom{x}{5}=\binom{y}{2}$ has exactly $20$ integer solutions:

Y. Bugeaud, M. Mignotte, S. Siksek, M. Stoll, Sz. Tengely, Integral Points on Hyperelliptic Curves, arXiv

I don't know if this can be proven by elementary means. And I don't know the situation for $\binom{x}{5}=2 \binom{y}{5}$. I just want to warn you that it might be a waste of time to look for elementary solutions, and that instead more sophisticated methods are necessary. On the other hand, this equation arises as a problem from a book, so I am not sure ...

I am putting some results that may or may not be part of an answer here, as a community wiki post, rather that cluttering the question with them. Perhaps they will help lead someone to a complete answer. If you have similar potentially useful information or partial answers, please feel free to add it here.

Let's see what can be gleaned by looking at $\binom{n}{r} = 2 \binom{n-k}{r}$ for other values of $r$. There is an interesting duality: $\binom{n}{r} = 2 \binom{n-k}{r} \iff \binom{n}{k} = 2 \binom{n-r}{k}$. So we can find solutions to the original problem by finding solutions to $\binom{n}{k} = 2\binom{n-5}{k}$. For any $r$, there is a "standard" solution, $\binom{2r}{r} = 2\binom{2r-1}{r}$, and several "trivial" solutions, $\binom{i}{r} = 2\binom{j}{r}$ whenever $0 \leq i,j \leq r - 1$.

For $r = 1$, there are infinitely many solutions; for any $k$, we have $\binom{2k}{1} = 2 \binom{k}{1}$. Under the duality, these correspond to the "standard" solutions $\binom{2k}{k} = 2 \binom{2k - 1}{k}$.

For $r = 2$, there are also infinitely many solutions. It's fun to see why. A solution satisfies the equation $n(n-1) = 2 (n-k)(n-k-1)$ or $$\begin{equation}\tag{1}\label{eq:qud}n^2 - (4k+1)n + (2k^2 + 2k) = 0.\end{equation}$$ Thus $n = \frac{4k + 1 \pm \sqrt{8k^2 + 1}}{2}$. Since $8k^2 + 1$ is odd, this either has two integer solutions if $8k^2 + 1$ is a perfect square, or none at all. Thus, whenever there is one solution $n$ for a given difference $k$, there must be a second. The second key ingredient is that since we are multiplying evenly many terms, we can change the signs of all the terms without changing the result.

So, start with the standard solution: $$ 4 \cdot 3 = 2 (3 \cdot 2). $$ Then it's also true that $$ 4 \cdot 3 = 2 (-2 \cdot -3), $$ in other words, $[4]_2 = 2[-2]_2 = 2[4-6]_2$, a solution with $k = 6$. So there must be another. When $k = 6$, equation $\eqref{eq:qud}$ is $n^2 - 25n + 84 = 0$, and we know we can factor out $(n - 4)$, which leaves $(n - 21)$. And indeed, $\binom{21}{2} = 2 \binom{15}{2}$. The duality gives $\binom{21}{6} = 2 \binom{19}{6}$. Repeating the process, we get $[21]_2 = 2 [-14]_2$ with a difference $k = 35$; factoring $(n - 21)$ from equation $\eqref{eq:qud}$ leaves $(n - 120)$, and indeed $\binom{120}{2} = 2 \binom{85}{2}$. Dually, $\binom{120}{35} = 2 \binom{118}{35}$. We can keep going up this "staircase" forever; if $k$ gives a solution, then so does $3k + \sqrt{8k^2 + 1}$. This is OEIS A001109.

Of course, this doesn't help for our case because $5$ doesn't appear. But it does show that solutions exist, besides the standard and trivial ones, for $r > 2$.

Now consider $r = 3$. We need to solve $[n]_3 = 2[n-k]_3$ or $$\begin{equation}\tag{2}\label{eq:cub}n^3 - (6k + 3)n^2 + (6k^2 + 12k + 2)n - (2k^3 + 6k^2 + 4k) = 0.\end{equation}$$ With $k = 0$ this factors as expected as $n(n-1)(n-2)$. With $k = 1$ it again gives us the known trivial and standard solutions, $(n-1)(n-2)(n-6)$. With $k = 2$, $\eqref{eq:cub}$ is $n^3 - 15n^2 + 50n - 48 = 0$, from which we can factor the known trivial solution: $(n-2)(n^2 - 13n + 24) = 0$. But the other factor has no integer solutions. For general $k$, we can apply the Cubic formula. We find that the discriminant $\Delta$ is $4(-27k^6 + 108k^4 + 18k^2 + 1)$, which is positive for $k = 0,1,2$ and negative for $k \geq 3$, so that it has three real roots for $k \leq 2$ but only one real root for $k \geq 3$. This root turns out to be $$ 2k + 1 - \sqrt[3]{-3k^3 + \sqrt{k^6 - 4k^4 - \frac{2}{3}k^2 - \frac{1}{27}}} - \sqrt[3]{-3k^3 - \sqrt{k^6 - 4k^4 - \frac{2}{3}k^2 - \frac{1}{27}}}. $$ I don't think that this can ever be an integer, but I don't know how to prove this. Unfortunately it's not sufficient to show that the contents of the cube roots cannot be perfect cubes, since even if neither is an integer, their sum still could be. However, no solutions exist for $n$ up to 10,000 (checked in the table provided at OEIS A000292.)

We can at least check if any answers to our original question have the difference $n - m = 3$ by looking at the above root, with $k = 5$. It's about 25.268, not an integer, so any extra solutions have the difference at least 4.

For $r = 4$, a similar "negation" trick should work as for $r = 2$. Unfortunately, it doesn't seem to yield extra positive solutions. The equation $[n]_4 = 2[n-k]_4$ expands to $$\begin{equation}\tag{3}\label{eq:qrt}n^4 - (8k + 6)n^3 + (12k^2 + 36k + 11)n^2 - (8k^3 + 36k^2 + 44k + 6)n + (2k^4 + 12k^3 + 22k^2 + 12k) = 0.\end{equation}$$ Starting from the standard solution, $[8]_4 = 2[7]_4$, we have another solution, $[8]_4 = 2[-4]_4$, with difference $k = 12$. And indeed, we can factor $(n - 8)$ out of equation $\eqref{eq:qrt}$ with $k = 12$, $$ n^4 - 102n^3 + 2171n^2 - 19542n + 65520 = (n-8)(n^3 - 94n^2 + 1419n - 8190). $$ But the other factor is a cubic with one real root, which isn't an integer. There are no solutions (besides the trivial and standard) up to $n=1002$ (checked in the table provided at OEIS A000332.)

When $k=5$, $\eqref{eq:qrt}$ becomes $n^4 - 46n^3 + 491n^2 - 2126n + 3360$, which (according to Wolfram Alpha) has no integer solutions. So for the original problem, $n - m > 4$.

Related OEIS sequences:

• A000389: $\binom{n}{5}$ and table up to $n=1000$
• A001109 are values of $k$ such that $\binom{n}{2} = 2\binom{n-k}{2}$, or equivalently $\binom{n}{k} = 2\binom{n-2}{k}$, has a solution $n = \bigl(4k + 1 + \sqrt{8k^2 + 1}\bigr)/2$.
• A082291 are the values of $n-2$ in the above, offset by one. That is, A001109 begins 0,1,6 and A082291 begins 2, 19, 118. $\binom{2+2}{1} = 2\binom{2}{1}$, and $\binom{19+2}{6} = 2\binom{19}{6}$.