Azerbaijan Problem-1

Consider two cases -

Case 1 : $a=b$

In this case, the given condition tells that $(a!)^2$ is divisible by $2a!$ . That is $a!$ is divisible by $2$ . The minimum value of $a$ is $2$ to satisfy this condition, which is also the value of $b$ . Then the value of $3a-2b$ is $\boxed 2$ .

Case 2 : $a\neq b$

In this case, let $c$ be a positive integer. For $a=b+c, 3a-2b=b+3c$ . For $a=b-c, 3a-2b=b-3c<b+3c$ . So to attain the minimum of $3a-2b$ , take $b=a+c$ , and again to achieve the minimum, take $c=1$ such that $b=a+1$ . Then we have the condition $a! (a+1)!$ is divisible by $a! +(a+1)!=a! (a+2)\implies (a+1)!$ is divisible by $a+2$ . The minimum value of $a$ to satisfy this condition is $4$ . So $b=5$ and $3a-2b=3\times 4-2\times 5=\boxed 2$ .

The main sketch of a proof is given here:

1- A solution is found ( $3a-2b=2$ )

2-we show $3a-2b$ cannot be less than $0$

3- finally we need to prove $3a-2b$ cannot be equal to $1$

1- first part

if $a=b$ , then

$2(a!)|a!a! \implies 2|a!$

for any $a \geq 2$ , $(a,a)$ is a solution for the relation. The minimum would be $3a-2a=a=2$ . Note that $(a,b)=(4,5)$ is a solution as well (it gives the same minimum $2$ ).

2- second part

If $a>b$ , one can manipulate

$(a!+b!)|a!b! \implies b!(\frac{a!}{b!}+1)|a!b! \implies \frac{a!}{b!}+1|a!$

Assume $\frac{a!}{b!}+1|a!$ is true. If, for a prime $p$ , $p|\frac{a!}{b!}+1$ , then $p \nmid \frac{a!}{b!}$ , which means $p|b!$ . Consequently

$\frac{a!}{b!}+1|b! \implies \binom{a}{b} (a-b)!+1|b! \implies \binom{a}{b} (a-b)! < b!$

since $a\neq b$ , we have $\binom{a}{b}>2$ and $a-b$ must be less than $b$ in order for $\binom{a}{b} (a-b)! < b!$ to be true.

$a-b<b \implies a<2b$

The problem is symmetrical with respect to the relation, but not the term $3a-2b$ . if $a=b+d, \ d\in \mathbb{N}$

$3a-2b=3a-2(a-d)=a+2d>0$

and if $b=a+d, \ d\in \mathbb{N}$ , knowing $a<2b$

$3a-2b=3a-2(a+d)=a-2d>0$

Therefore, $3a-2b$ is always greater than $0$ .

3- last part

if $3a-2b=1$ then, the solutions should be of form $a=1+2n$ and $b=1+3n$ , for $n\in \mathbb{N}$ . Although difficult, but it can be proven, by induction over $n$ , that such solutions $a=1+2n$ and $b=1+3n$ cannot satisfy the relation $a!+b!|a!b!$