Proof that $n!$ is divisible by $(n+1)^2$ for $n=xy+x+y$ I noticed the other day that for $n&gt;8, n\in\mathbb{N}$, the factorial $n!$ seems to be divisible by $(n+1)^2$ when $n$ can be written in the form $xy+x+y$ (where $x,y\geq1$ and $\in\mathbb{N}$). Some examples:


$n=14=2 \times 4+2+4$ (i.e. $x=2$, $y=4$), and we have $14!|15^2$
$n=15=3 \times 3+3+3$ (i.e. $x=y=3$), and we have $15!|16^2$,
$n=19=3 \times 4+3+4$ (i.e. $x=3$, $y=4$), and we have $19!|20^2$,
...


This seems to hold for the first $15$ values I checked, so it seemed natural to try to prove it for all $n=xy+x+y$, however I could not find a proof. Here is my attempt:

Plugging $xy+x+y$ into the expressions above and noticing that $xy+x+y+1=(x+1)(y+1)$, we need to prove that

$$
(xy+x+y)!|(x+1)^2(y+1)^2,
$$

or

$$
\Gamma((x+1)(y+1))|(x+1)^2(y+1)^2.
$$

It is easy to prove that 

$$
\Gamma((x+1)(y+1))|(x+1)(y+1)
$$

however I am struggling to prove the same for $(x+1)^2(y+1)^2$. Any hints?
Wlog. $x\le y$.
If we are lucky, we find $x+1,2(x+1),y+1,2(y+1)$ as different elements among the factors $1,2,3,\ldots, n$ and we are done.

How can we fail to be lucky? The four candidates may be too large or not be different after all. As we assume $x\le y$, the following list is exhaustive:


It may happen that $2(y+1)&gt;n=xy+x+y$. This implies $x=1$. And indeed, $x=1$, $y=2$ gives $n=5$ and $36$ does not divide $5!$. More generally if $y+1$ is prime, then $(2y+1)!$ is not divisible by $(y+1)^2$.
It may happen that $x+1=y+1$, which of course means that $x=y$. If $x\le 3$, we have $n\le 8$, which is excluded. If $x\ge 4$,  we can use the factors $x+1,2(x+1),3(x+1),4(x+1)$ instead. These are guaranteed to be different and we have $4(x+1)\le xy+y&lt;n$.
It may happen that $2(x+1)=y+1$. Then we can use $x+1,2(x+1)=y+1, 3(x+1), 2(y+1)$ instead.


Thus


  If $n=xy+x+y&gt;8$ with positive integers $x,y$ then $(n+1)^2\mid n!$ except when $n+1$ is twice a prime.
$\color{Green}{\text{Question}}$ : 
Find all integers $n$; 
such that $\color{Green}{(n+1)! \mid n!} $ . 





$\color{Blue}{\star \ \ \ \ \text{condition}}$ : 

Now suppose that $n+1$ can be written 
in the form $n+1=ab$; 
with $2 &lt; a,b$. 
Also assume that one of the following occures: 


If $a= b$; then $5 \leq a=b$.
If $a=2b$; then $5 \leq b  $.
If $2a=b$; then $5 \leq a  $.




With the condition as in 
$\color{Blue}{\star}$ ; 
we claim that 
the assertion is $\color{Blue}{\text{true}}.$ 



Let $n, a, b$ to be fixed as above in $\color{Blue}{\star}$ . 
Notice that if we let $x:=a-1$ and $y:=b-1$ 
then one can easilly see that: 
$$n=xy+x+y.$$ 





$\color{Blue}{\text{Remark(I)}}$ : 
With the notation as in $\color{Blue}{\star}$; 


If $a=b$ , with $5 \leq a$ 
then we have: 
$$b &lt; 2b &lt; 3b &lt; 4b &lt; ab=n+1 \Longrightarrow b &lt; 2b &lt; 3b &lt;4b \leq n=ab-1 \ ; $$ 
so we can conclede that: 
$$(b)(2b)(3b)(4b)=24(ab)^2=24(n+1)^2 \mid n! ;$$ 
which implies that: $(n+1)^2 \mid n! \ \ $ . 





If $a=2b$ ;  with $5 \leq b$ 
then we have: 
$$b &lt; 2b &lt; 3b &lt; 4b &lt; ab=n+1 \Longrightarrow b &lt; 2b &lt; 3b &lt;4b \leq n=ab-1 \ ; $$ 
so we can conclede that: 
$$(b)(2b)(3b)(4b)=6(ab)^2=6(n+1)^2 \mid n! ;$$ 
which implies that: $(n+1)^2 \mid n! \ \ $ . 





If $2a=b$ ;  with $5 \leq a$ 
then we have: 
$$a &lt; 2a &lt; 3a &lt; 4a &lt; ba=n+1 \Longrightarrow a &lt; 2a &lt; 3a &lt;4a \leq n=ba-1 \ ; $$ 
so we can conclede that: 
$$(a)(2a)(3a)(4a)=6(ab)^2=6(n+1)^2 \mid n! ;$$ 
which implies that: $(n+1)^2 \mid n! \ \ $ .  





If $a \neq b$ and $a \neq 2b$ and $2a \neq b$ ; 
then we have: 
$$b &lt; 2b &lt; ab=n+1 \Longrightarrow b &lt; 2b \leq n=ab-1 \ ; $$ 
$$a &lt; 2a &lt; ab=n+1 \Longrightarrow a &lt; 2a \leq n=ab-1 \ ; $$
so we can conclede that: 
$$(b)(2b)(a)(2a)=4(ab)^2=4(n+1)^2 \mid n! ;$$ 
which implies that: $(n+1)^2 \mid n! \ \ $ .  














Remark(II): Integer $N$ 
can not be written 
in the form $N=ab$; 
with $2 &lt; a,b$; 
if and only if 
$N$ is prime or twice a prime. 



Remark(III): 
We can split the integres in three cases:


$\color{Red}{   n+1 =   \text{prime numbers and twice of prime numbers; 
for which the assertion is wrong!}   }.$ 
$\color{Purple}{ n+1= a.a 
 \ \ \ 
\text{for}    
\ \ \ 
a \in \{ 1, 2, 3, 4 \} }.$ 
This case has only four numbers; 
which can you check out that 
the assertion is $\color{Blue}{\text{true}}$ only for 
$\color{Blue}{n+1=16}$ and 
$\color{Blue}{n+1= 1}$ . 
$\color{Purple}{ n+1= a.(2a) 
 \ \ \ 
\text{for}    
\ \ \ 
a \in \{ 1, 2, 3, 4 \} }.$ 
This case has only four numbers; 
which can you check out that 
the assertion is $\color{Blue}{\text{true}}$ only for 
$\color{Blue}{n+1=18}$ and 
$\color{Blue}{n+1=32}$ . 




$\color{Green}{\text{Remark(IV)}}$ : 
Except for the following cases; 
for which the assertion is 
$\color{Red}{\text{wrong}}$ 

1.$\color{Red}{   n+1 =   \text{prime numbers and twice of prime numbers 
}   }.$ 

2.$\color{Red}{ n+1= 4, 9, 2, 8 } .$ 

for all the other cases 
the assertion is 
$\color{Blue}{\text{true}}$ . 













$\color{Red}{\text{Remark(V)}}$ : 
Let $p$ to be an odd prime number; 
then we have: 
$p^2 \color{Red}{\nmid} (2p-1)! \ \ . $ 



For example let $p$ to be an odd prime number; 
and let's define: 
$$n:=2p-1 
\ \ \ 
\text{and} 
\ \ \ 
x:=1 
\ \ \ 
\text{and} 
\ \ \ 
y:=p-1.$$

Notice that we have: $n=xy+x+y$



We calaim that $(2p)^2 \color{Red}{\nmid} (2p-1)!$. 

[ 
Suppose on contrary 
that $(2p)^2 \mid (2p-1)!$ , 
on the otherhand notice that 
$p^2 \mid (2p)^2$. 
So we must have 
$p^2 \mid (2p-1)!$ 
; 
which has an obvious contradiction 
with the above $\color{Red}{\text{Remark(V)}}$. 
]