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.
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.