How find this $x,y$ such this equation $x^x+x=y!$ 
Find all pairs of positive integers $(x,y)$,such
  $$x^x+x=y!$$

I find the $$(x,y)=(1,2)$$ is such it. and 
$$(x,y)=(2,3)$$
I think this equation have other roots. and maybe use inequality to solve this?
since Use  Bernoulli inequality we  have 
$$x^x=(x-1+1)^x\ge 1+x(x-1)=x^2-x+1$$
Thank you for you help, this problem is from Mathematical contest in china jiangxi province in 2014
 A: If $x>2$ is even, then $(x^{x-1}+1)x=y!$ shows that any power of $2$ dividing $y!$ divides $x$. Therefore $$x\geq 2^{\lfloor y/2 \rfloor}.$$ For $x>2$ this implies $x>y$, a contradiction because $x^x+x>x!$.
If $x>1$ is odd and $x^x+x=y!$, then $y\geq 4$. Therefore $4 \mid y! = x^x+x$ and $$x^{x-1}\equiv -1 \bmod 4.$$
This implies that $-1$ is a quadratic residue $\bmod\, 4$, a contradiction.
A: There are no other such pair of integers for $x>2$.
Updated Proof:


*

*If $x$ is not prime:


*

*$x^x \geq x! \implies x^x+x>x!$

*Therefore, we must choose $y>x$

*$x^x+x$ is divisible by $x$ exactly once, because $\frac{x^x+x}{x}=x^{x-1}+1$ is not divisible by $x$

*But since $x$ is not prime, for every $y>x$ that we choose, $y!$ is divisible by $x$ at least twice


*If $x$ is a prime such that $x \equiv 1 \bmod 6$, then let's analyze $x^{x-1}+1$:


*

*$x^K \equiv 1 \bmod 6$ for every non-negative integer $K$

*$x^{x-1} \equiv 1 \bmod 6$, since $x-1$ is a non-negative integer

*$x^{x-1}+1 \equiv 2 \bmod 6$

*$x^{x-1}+1$ is not divisible by $3$

*$x^x+x$ is not divisible by $3$, hence, no $y>2$ exists such that $x^x+x=y!$


*If $x$ is a prime such that $x \equiv 5 \bmod 6$, then let's analyze $x^{x-1}+1$:


*

*$x^K \equiv 1 \bmod 6$ for every non-negative even integer $K$

*$x^{x-1} \equiv 1 \bmod 6$, since $x-1$ is a non-negative even integer

*$x^{x-1}+1 \equiv 2 \bmod 6$

*$x^{x-1}+1$ is not divisible by $3$

*$x^x+x$ is not divisible by $3$, hence, no $y>2$ exists such that $x^x+x=y!$


*Finally, if $x=3$ then $x^x+x=30$, and no $y$ exists such that $y!=30$
A: $y\le 1$ leads to no solution.
For $y>1$ we have $y!=\prod_{k=1}^y k<\prod_{k=1}^y y=y^y$.
Hence $y!=x^x+x$ implies $y^y>x^x$ and thus $y>x$ (the function $f(t)=t^t$ is obviously strictly increasing because $(t+1)^{t+1}>(t+1)^t>t^t$).
As you have already found the solution $x=1, y=2$, we may assume that $x>1$.
Let $p$ be a prime divisor of $x$. Then $x^{x-1}+1$ is not a multiple of $p$, hence neither is $\frac{y!}{x}$. That is, if we strike $x$ from the list $1, 2,\ldots, \not x,\ldots , y$, none of the remaining numbers is a multiple of $p$.
This leaves only the possibility that $x=p$ as otherwise $p$ remains on the list.
Thus $x$ is prime.
We already know the solution $x=2$, $y=3$, hence may assume that $x$ is an odd prime.
Then $x^{x-1}+1\equiv 2\pmod 4$, hence $\frac {y!}x$ is not a multiple of $4$. This implies $y<4$ and you have already found all solutions with $y<4$ by exaustive search.
A: It's easy to prove that if $x>3$ then $x<y,$ thus $x-1\mid y!,$ but $x^x+x\equiv 1^x+1\equiv 2\pmod{x-1},$ a contradiction.
