How many solutions for $(6b)^b\equiv (12b-k)^b\mod p$? Rephrasing my previous question;
If $(6b)^b\equiv (12b-k)^b\mod p$, where $b$ is odd and $p=1+6qb$, and where $p$ and $q$ are prime, are there any solutions for $k$ other than $k\equiv6b\mod p$?
 A: If I can offer some advice: if there's no reason to think something like this to be true, then you should expect there to be eventual counterexamples. In the future, when you have a question like this, I propose that it would be more useful for you (and everyone else) if you explain why you came up with this question and why you expected the answer you did, so that people can try to address the mathematics of the issue, instead of just getting a computer to find counterexamples for you.

Let the prime $q=7$ and let $b=3$, so that $p=1+(6\cdot 7\cdot 3)=127$. Then
$$(6\cdot 3)^3\equiv 117 \equiv (7\cdot 3)^3\bmod 127$$
thus demonstrating that $k\equiv 15\bmod 127$ is a solution other than $k\equiv 6b\equiv 18\bmod 127$.

Let the prime $q=7$ and let $b=5$, so that $p=1+(6\cdot 7\cdot 5)=211$. Then
$$(6\cdot 5)^5\equiv 185\equiv (4\cdot 5)^5\bmod 211$$
thus demonstrating that $k\equiv 40\bmod 211$ is a solution other than $k\equiv 6b\equiv 30\bmod 211$.

Let the prime $q=23$ and let $b=7$, so that $p=1+(6\cdot 23\cdot 7)=967$. Then
$$(6\cdot 7)^7\equiv 196\equiv(20\cdot 7)^7\bmod 967$$
thus demonstrating that $k\equiv 911\bmod 967$ is a solution other than $k\equiv 6b\equiv 42\bmod 967$.

Here was the Mathematica code I used (not intended to be even close to exhaustively finding all counterexamples below a given bound, as I said in a comment below; this was just a "first pass", which already turned out to be sufficient):

mylist = {{"p", "r", "q", "b"}};
Do[
   Do[p = 1 + 6*Prime[n]*x;
      Do[If[PrimeQ[p] && PowerMod[6, b, p] == PowerMod[r, b, p], 
         AppendTo[mylist, {p, r, Prime[n], b}]],
      {r, 1, Prime[n]}],
  {b, 3, Prime[n], 2}],
{n, 1, 10}];
TableForm[mylist]

