TITLE: How do I find solutions to $2^n+11 \equiv 0 \pmod n$?
QUESTION [8 upvotes]: In recent days, I have been studying the equation $m^n+h \equiv 0 \pmod n$ where $m,n \in \mathbb N$ and $h\in\mathbb Z $, and I have noticed that $2^n+11 \equiv 0 \pmod n$ have no solutions in $1 \leq n \leq 2000000000$ except for $1$ and $13$. How can I find other solutions that satisfies the equation, or prove otherwise that no solutions $>13$ exists?

REPLY [4 votes]: $n=383979411456776027$
Let valid prime $p$ such that exist $k$ for $2^k\equiv -11\pmod{p}$. For brute force need pick up set triples $(p,k,h)$, where $h=ord_p(2)$. Then $n=p(k+j\cdot h)$, where $j$ is brute force step. For speed up calculation can use CRT of two valid triples.
gp-code:
 P= read("n11.dbt");
 for(i=2, #P~, for(j=1, i-1,
  c= iferr(chinese(Mod(P[i,1]*P[i,2], P[i,1]*P[i,3]), Mod(P[j,1]*P[j,2], P[j,1]*P[j,3])), Err, 0);
  if(c,
   k= lift(c); h= c.mod;
   d= 10^10\h; 
   for(t=d, d+10^4,
    n= k+t*h; \\print(h"    "n);
    if(Mod(2,n)^n==-11,
     print(n"    "k"    "h"    "t)
    )
   )
  )
 ))

File "n11.dbt" contains valid triples:
[13, 1, 12; 23, 10, 11; 29, 11, 28; 43, 5, 14; 47, 17, 23; 71, 11, 35; 83, 65, 82; 89, 8, 11; 97, 35, 48; 101, 63, 100; ...]. For $p<10^7$ i pickuped 180561 triples, but still they have many non-valid  triples, becose for me algorithm selecting triples is not simple.