show this number is not a prime number The following problem is a special case Show this number always is composite number? , but I think this special case is relatively difficult to deal with, that is, if this is solved, it may solve the general situation.  

show this number
  $$A=2004^{2005}+1002^{2005}\cdot 2005^{1002}+2005^{2004}$$is not a prime number.

This is  a question from an eighth-grade math contest exercise, (this .：see:at last page problem 20
I'm thinking mod 3, mod 5, mod 7. It doesn't seem to work.
because
$$A\equiv 0+0+1=1\pmod 3$$
$$A\equiv -1+0+0=4\pmod 5$$
Use Fermat's little theorem
$$A\equiv 2+1+3^{0}=4\pmod 7$$
 A: Actually the smallest prime factor of A turns out to be $670543$. This limits the number of approaches that could possibly work: don't bother checking mod small primes.
A: $$2004^{2005}+1002^{2005} 2005^{1002}+2005^{2004}$$
$$x=2004$$
$$x^{x+1}+\left(\frac{x}{2}\right)^{x+1} (x+1)^{x/2}+(x+1)^x=2^{-x-1} \left(x^{x+1} (x+1)^{x/2}+2^{x+1} x^{x+1}+2^{x+1} (x+1)^x\right)$$
which should never be prime when $x$ is even:
Consider the portion (ignoring the extra $2^{x+1}$ for now):
$$\left(x^{x+1}+(x+1)^{x}\right)$$
Then, let $x={1,2,3,...}$, but don't evaluate to see:
$$x^2+x+1$$
$$x^3+x^2+2 x+1$$
$$x^4+x^3+3 x^2+3 x+1$$
...
Then the other portion: 
$$\left(x^{x+1}(x+1)^{x/2}\right)$$
$$x^4+x^3$$
$$x^7+2 x^6+x^5$$
$$x^{10}+3x^9+3x^8+x^7$$
...
So the inside part of the polynomial is the sum of two different variations of binomial expansions with different powers for $x$, so it cannot be prime when $x$ is even.
One other note: it's almost like $2^2(x^2+x+1)+x^4+x^3$ characterizes the entire function.
I'm pretty sleepy right now, if I missed something let me know. (nicely, please)
