<s>[INST] 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
 [/INST] 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 $y!=30$
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