Proving $10240...02401$ composite

I got this question recently, and have been unable to solve it.

Prove that $1024\underbrace{00 \ldots\ldots 00}_{2014 \text{ times}}2401$ is composite.

I have two different ways in mind.

First is $7^4+400(2^2\cdot10^{504})^4$, which looks like Sophie Germain, but I'm not sure what to do with the $400$. Another thought is that this is almost a palindrome, with the order of just two digits interchanged. I'm not sure where to go from there, and if it'd provide any results, but it seems interesting nonetheless.

Please help.

Solution 1:

The method introduced in this modification is for reduction of volume of calculations.

Let's start with a simple example; consider number $N=1024002401=1670477\times 613$. We may write:

$N=2^{10}.10^6+24.10^2 +1=2^3.10^2(2^7.10^4+3)+1$

$2^7.10^4+3=1280003 ≡59 \mod (613)$

⇒ $N=2^3.10^2(613 k + 59)+1=800\times 613\times k +800 \times 59 +1$

Let $800\times 613 =a$ and $800\times 59 +1 =b$

⇒ $(a, b)=613$

Now suppose we do not know that $p=613$ and $r=59$ in following relation:

$(a, b)=(800 p, 800 r+1)=p$

$a=800 r +1$ gives infinite numbers which its factors may be candidates for a factor of a number like ($N=2^{\alpha}.10^{\beta}+3^{\gamma}$).

For example giving r numerous values we find that for $r=59$ we get $800\times 59+1=47201=7\times 11\times 613$. Each of theses three factors can be the candidate for p to be tried to find the factor of N. For $N=2^{10}.10^{2018}+7^4$ we may write:

$N=2^{10}.10^{2018}+7^4=2^3.10^2(2^7.10^{2016}+3)+1$

$2^7.10^{2016}+3≡ r\ mod (p)= k.p +r$

$N=800(k.p+r)+1=800p.k+800r +1$

Now values for $r$ in $(800 r +1)$ give numbers their factors can be checked as a primes factor of N.