Prove that sum of 2 double factorials equals to a factorial only on limited solutions

Question

Suppose a! = b!! + c!!, prove that a + b + c < 2022

I can only prove that both b, c are odd numbers.

I wrote a program to search for solutions, which only gives (2,1,1), (3,3,3) and (5,5,7). But I can't prove that they are the only solutions.

Answer 1

Let's prove we know all solutions. Here is an outline of a proof strategy. Since I shall have not the time to write down the details in the near future, I hope the steps indicated are sufficiently clear so as their proofs present no difficulties.

Below we suppose $1\le b\le c$ and $bc$ odd.

Fact 1. The only solutions with $b=c$ are $(a,b,c)=(2,1,1), (3,3,3)$.

Fact 2. The only solution with $b=1$ is $(a,b,c)=(2,1,1)$.

From now on we may suppose $3\le b<c$. For each such solution, we shall put $R=b!!+c!!$.

Fact 3. There exist positive integers $s$, $t$ such that $\{b,c\} = \{8s\pm 1, 8t+5\} $. In particular, $b$, $c\in \{5, 7, 9, 13, 15, 17,21, 23, 25, 29, \ldots\}$.

Fact 4. $b=5$ only for $(a,b,c)=(5,5,7)$.

Fact 5. No solution exists with $7\le b<53$

Fact 6. (Legendre's formula) $$v_p(x!)= \lfloor \frac{x}{p} \rfloor + \lfloor \frac{x}{p^2} \rfloor +\lfloor \frac{x}{p^3} \rfloor + \cdots$$

Standing hypotheses $53\le b<c, \quad 34 \le a< c . $

For an integer $x\ge 2$, let $q(x)$ denote the largest prime which is less than or equal to $x$.

Fact 7. (Nagura's analogue of Bertrand's postulate) There exists a prime between $x$ and $1.2x$ for any $x\ge 25$.

J. Nagura, On the interval containing at least one prime number, Proc. Japan Acad., Ser. A 28, no. 4 (1952), 177--181. doi:10.3792/pja/1195570997.

Fact 8. $q(a)=q(b)$.

Fact 9. $5b+1 \le 6a$

Fact 10. $c-b\ge 6$, $v_3(b!! +c!!)=v_3(b!!)$.

Fact 10. $v_3(a!)>v_3(b!!)$.

From the last two facts one concludes that the equation has no solution satisfying Standing hypotheses.