Why y>2008?

Because $ 21^y>(21n)^{2008}>21^{2008}$.

Ok, I did another thing! Nice solution! Congratulations

Another solution: First see that $ 21 \mid x$, then $ 3^{1000} \mid 2008!$ so $ y \ge 1000$ but then $ 21^{1000} \mid 2008!$ contradiction.

2009 is prime so by fermat and wilson Lhs divisible by 2009. but 2009 does not divide \[ \ 21^y \]

2009 is not a prime Please check before posting $ 2009=7^2*41$ Daniel

x^2008+2008!=21^y 2008!=0 mod(21) => x^2008=0 mod(21) => x=0 mod(21) => x=21k (21k)^2008+2008!=21^y 2008!= 21^y-(21k)^2008, when 21^y-(21k)^2008= 2008!>0 => 21^y>(21k)^2008 2008!= 21^2008(21^(y-2008) - k^2008) => 21^2008\ 2008! =><=

Another Solution : Note $3^y||2008!,7^y||2008!$ It's obviously impossible so no solution. Generalization : Solve $x^n+n!=(p_1p_2...p_k)^y$ where $p_1,p_2......,p_k$ are some odd primes less than $n$ and $k>1$.

I thought my solution was nice but when i started reading the comments above i realized i did some unnecessary stuff $x^{2008} + 2008!=21^y$ $x^{2008} +0 \equiv 0 (mod 21) \implies 21| x , x=21l$ for some integer $l$ which means the equation becomes: $2008!=21^y-(21l)^{2008}$ $\implies y>2008$ Now notice that $\lfloor{\frac{2008}{7}}\rfloor+ \lfloor\frac{2008}{49}\rfloor + \lfloor\frac{2008}{343}\rfloor<500$ so $2008!$ can not have more than $500$ factors of $21$(it is easy to check that it has more factors of $3$ than $7$). $2008!=21^y-(21l)^{2008}$ $\frac{2008}{21^{500}}=21^{y-500}-21^{1708}l^{2008}$ but the $LHS$ is not an integer. $\square$