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 $$A\equiv 2+1+3^{0}=4\pmod 7$$

  • $\begingroup$ As an extension, it might also be interesting to find for which $x$ makes $$(2x)^{2x+1}+x^{2x+1}\cdot (2x+1)^x + (2x+1)^{2x}$$ prime... $\endgroup$ – TheSimpliFire May 17 '18 at 18:45
  • $\begingroup$ @TheSimpliFire That is this question exactly: math.stackexchange.com/questions/2771998/… $\endgroup$ – Sil May 17 '18 at 19:24
  • $\begingroup$ @Sil Thanks! I'll have a look at it. $\endgroup$ – TheSimpliFire May 19 '18 at 15:08

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.

  • 1
    $\begingroup$ Let $x=1002^{2005},\ y=2005^{1002}, then\\ A=x^2+xy+y^2$ $\endgroup$ – herb steinberg May 9 '18 at 3:15
  • $\begingroup$ @herbsteinberg No, look again. But even if it were true, it wouldn't help. $\endgroup$ – Robert Israel May 9 '18 at 3:17
  • $\begingroup$ @herbsteinberg $x^2 \neq 2004^{2005}$. $\endgroup$ – N. S. May 9 '18 at 3:17
  • $\begingroup$ You are right - it should be$2^{2005}x+xy+y^2$ $\endgroup$ – herb steinberg May 9 '18 at 3:22
  • $\begingroup$ Explain, please, (my English is quite poor) what you mean by "smallest prime factor of A turns out to be 670543". Does it mean that this number is in effect a factor of A? In this case you have shown that A is not a prime and then you would deserve the recognition of the best answer. $\endgroup$ – Piquito May 10 '18 at 14:32

$$2004^{2005}+1002^{2005} 2005^{1002}+2005^{2004}$$ $$x=2004$$ $$x^{x+1}+\left(\frac{x}{2}\right)^{x+1} (x+1)^{x/2}+(x+1)^x=2^{-x-1} \left(x^{x+1} (x+1)^{x/2}+2^{x+1} x^{x+1}+2^{x+1} (x+1)^x\right)$$

which should never be prime when $x$ is even:

Consider the portion (ignoring the extra $2^{x+1}$ for now):


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:


$$x^4+x^3$$ $$x^7+2 x^6+x^5$$ $$x^{10}+3x^9+3x^8+x^7$$ ...

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)


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