For any positive integer $n$ we define $n!!=\prod_{k=0}^{\lceil n/2\rceil -1}(n-2k)$. Prove that if the positive integers $a,b,c$ satisfy $a!=b!!+c!!$, then $b$ and $c$ are odd. Proposed by Mihai Cipu
Problem
Source: Stars of Mathematics 2024 P2 (senior level)
Tags: number theory, double factorial
sami1618
20.12.2024 01:31
What about $3!!=2!!+1!!$?
oVlad
20.12.2024 01:48
There's a mistake in the statement. It's supposed to say $a!=b!!+c!!$.
tabel
25.12.2024 01:54
It's pretty messy . If I've made any mistakes I'd be grateful if you would tell me .
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
Feedback would be nice =)