For $a=1$ there's obviously no solutions For $a\ge 2$ , LHS is even , which means $b!!$ and $c!!$ are congruent mod 2 , respectively $b\equiv c$(mod 2) Suppose by contradiction , that there exists $b$ and $c$ , both even , satisfying the given condition . $b=2x,c=2y $ with $x,y\in \mathbb{N}^*$ $\implies b!!=2^{x}x!$ $c!!=2^{y}y!$ Suppose WLOG $y\ge x \implies y=x+n$ , $n\in \mathbb{N}^*$(for $n$=0 , it's trivial to prove that it leads to no solutions) Case 1 , $a$ is odd: $a=2t+1$ , $t\in \mathbb{N}^* \implies a! = (2t+1)!=(2t+1)!!2^{t}t!$ We have $(2t+1)!!=\frac{2^{x}x![1+2^{n}(x+1)(x+2)...(x+n)]}{2^{t}t!}.$ For $t<x$ : LHS is odd , while RHS is even - absurd . For $t>x$ : since LHS is an integer it means $2^{t-x}\frac{t!}{x!}|1+2^{n}(x+1)(x+2)...(x+n) \implies 2|1+2^{n}(x+1)(x+2)...(x+n) $, which is an odd integer - absurd. Thus , we get $x=t \implies (2x+1)!!=1+2^{n}(x+1)(x+2)...(x+n)$. For $n=1$ there are no solutions. For $n\ge 2$ : $(x+1)$ or $(x+2)$ is odd , and is among the terms in the product $(2x+1)!!$ . Taking modulo $x+1$ , respectively $x+2$ we get $0\equiv 1$ - absurd , yet again . This case yields no solutions . Case 2 ,$a$ is even : $a=2t$ , $t\in \mathbb{N} , t\ge2$($t=1,a=2$ leads to no solutions with b and c even) $a!=(2t)!=2^{t}t!(2t-1)!!$ , and then we proceed the same as in case 1 . Q E D