Find a composite number $n$ satisfies $(2+3I)^n≡2-3I\pmod{n}$ As we know if $p$ is an odd prime number then $$(a+bI)^p\equiv a+(-1)^\frac{p-1}2bI\pmod{p},$$
where $I=\sqrt{-1}$. However, is there any composite number $n$ that satisfies $$(2+3I)^n≡2-3I\pmod{n}\quad ?$$
I know $n$ is a solution to the equation $13^{n-1}\equiv 1\pmod{n}$, but cannot go on.
Many thanks in advance.
 A: $${\Large \textrm{There is no such }n\leq10^{10}}$$
When doing a brute-force numerical search on this problem, there are two obvious modes of attack.
One method is to compute $(2+3I)^n\ \forall\ n\leq N$ iteratively via $(2+3I)^n = (2+3I)(2+3I)^{n-1}$, and only reduce $\operatorname{mod} n$ to check the congruence for each $n$. $(2+3I)^n$ has $O(n)$ digits, so each multiply takes $O(n)$ time. Doing $N$ such multiplies brings the overall complexity to $O(N^2)$.
A more efficient algorithm is to compute $(2+3I)^n\ (\operatorname{mod}\ n)$ separately for each $n$ using e.g. a binary ladder, reducing $\operatorname{mod} n$ at each stage of the binary ladder.  Checking a single $n$ takes $\log(n)$ this with approach, resulting in a total time complexity of $O(N\log N)$ to check all $n\leq N$.
The complexity arguments given above are only estimates. However, I implemented both algorithms and found that they are borne out in practice: the two algorithms both take ~3s to check all $n\leq 10^5$ using one core on my system. From that point on, the first algorithm takes roughly quadratic time, whereas the second one takes roughly linear time.
Here is my implementation in PARI/GP of the faster of the two algorithms:
\\ Our search range
NMIN=2;
NMAX=100000000;


\\ Computes x^y (mod n) using a binary ladder
\\ This is Algorithm 9.3.1 from Crandall and Pomerance, "Prime Numbers", 2 Ed.
\\ RETURNS x (RATHER THAN 1) IF y=0...
pow_modN(x,y,n)={
  \\ Get: 1) y_binary (the binary expansion of y)
  \\      2) D        (the number of bits in y_binary)
  \\         Unfortunately these are not interleaved!
  y_binary = binary(y);
  y_shl_D=y;
  D=0;
  while(y_shl_D,
    D++;
    y_shl_D>>=1;
  );

  \\ Initialize to z=x.
  \\ Loop over bits of y, starting with the next-to-highest
  \\ y_binary[1] is the MSB, y_binary[D] is LSB.
  z=x;
  for(j=2, D,
    z*=z;
    if(y_binary[j], z*=x);
    z%=n;
  );

  \\ Return z=x^y (mod n)
  z;
}


\\ We work in the field of Gaussian integers, Z(w) here w==I==sqrt(-1).
w=quadgen(-4);

\\ We intend to exponentiate a.
a = 2+3*w;

\\ Compute a^n for all n in [NMIN,NMAX]
for(n=NMIN, NMAX, {
  zPowN_modN = pow_modN(a, n, n);
  re_zPowN_modN = real(zPowN_modN);
  im_zPowN_modN = imag(zPowN_modN);

  if(re_zPowN_modN-2,,
    if(im_zPowN_modN-n+3,,
      if(isprime(n),,
        print(n);
      );
    );
  );

  if(n%100000,,printf("Done to %u\n", n));

});

quit;


Negative results with numerical searches like this one are dangerous, because many different errors would produce the same result. For example, in C, (-1)%4 == 3%4 evaluates to -1==3 which is false, despite the fact that $-1\equiv3\ (\operatorname{mod}\ 4)$. To get the desired result, one has to check (-1-3)%4==0, which returns true.  Considering this, double checks are necessary.
For the above code I checked the result in two ways. Firstly, I dropped the requirement that $n$ be composite, which results in a long list of primes $n=3, 7, 11, 19, 23, \ldots$ I redid this calculation with a different algorithm (the "first algorithm above"), different software (Maple) and different hardware (my laptop), and compared the resulting lists up to $10^6$, with no differences found.
My second double check was to make a minor modification to the above code to check the congruence $(2+3I)^n \equiv 2+3I\ (\operatorname{mod}\ n)$ for composite $n$. This quickly finds many solutions:
1105,
2465,
10585,
29341,
41041,
46657,
115921,
162401,
$\ldots$, which agrees with the list given by Zev Chonoles in the comments.
A: I have managed to come up with anything conclusive, but here are some notes that hopefully can be inspiring.
1. I doubt this is well known as true, since if true it gives a fast definitive primality check for numbers of the form $4k+3$. Then for example the probable prime records wouldn't have to include any such numbers in the list for which "nobody knows how to prove or disprove ... primality." (I have already checked two of them.)
2. If $p\equiv 1 \pmod{4}$ then $(2+3i)^{p-1}\equiv 1 \pmod {p}$. If there is any $k<p-1$ such that $(2+3i)^k\equiv2-3i\pmod{p}$ then 
$$
(2+3i)^{k+1}\equiv 13\pmod{p}
$$
and by considering the norms
$$
13^k\equiv 13 \pmod{p} \\
(2+3i)^{(k+1)(k-1)}\equiv 1 \pmod{p}
$$
I'd like to generally rule this case out, but I cannot, since it holds for $k=19,p=61$.
If there's no such $k$, then $p$ cannot divide $n$, so it seems prudent to concentrate on numbers only divisible by primes that are 3 modulo 4.
3. If $p\equiv 3\pmod{4}$ then $(2+3i)^{p^2-1}\equiv 1 \pmod{p}$ and we might find a solution if $n\equiv p \pmod{p^2-1}$. In particular there will be a solution with $n=p^3$ if $13^{p-1}\equiv 1 \pmod{p^3}$. I wasn't able to find any such prime, but suspect it can't easily be ruled out e.g. with Hensel's lemma, since there are solutions to $13^{p-1}\equiv 1 \pmod{p^2}$ (e.g. $p=863$).
4. Generalizing from 3, it would be sufficient to find $n$ a product of primes that are 3 mod 4 such that for each prime $p$ dividing $n$
$$
\frac{n}{p} \equiv 1 \pmod{p^2-1}
$$
If $n$ is a product of distinct primes then they all must satisfy $p^3<n$ and hence there must be at least 4.  
