A congruence holding for primes $p \equiv 1,11 \pmod {12}$ Can you provide a proof or counterexample to the following claim ?

Let $p$ be prime number greater than three and let $T_n(x)$ be Chebyshev polynomial of the first kind , then $T_{p-1}(2) \equiv 1 \pmod p$ if and only if $p \equiv 1,11 \pmod {12}$  .

I have tested this claim up to $2 \cdot 10^6$ .
I was searching for a counterexample using the following two PARI/GP codes :
FermatChebyshev1(lb,ub)={
forprime(p=lb,ub,
if(Mod(p,12)==1 || Mod(p,12)==11 ,
if(!(Mod(polchebyshev(p-1,1,2),p)==1),print(p))))
}

FermatChebyshev2(lb,ub)={
forprime(p=lb,ub,
if(!(Mod(p,12)==1 || Mod(p,12)==11) ,
if(Mod(polchebyshev(p-1,1,2),p)==1,print(p))))
}

 A: There are no counter-examples.  It is a variation on Gauss's little lemma.
It relates to the short-chord function, which follows this 'isoseries' relation: $T_{n+1}=a T_{n}-T_{n-1}$.
It can be demonstrated that the isoseries as defined above, that steps of some x, correspond to an 'isopower' of a, ie $a_x$, in a way that $a_x$ itself forms an isopower.  The resulting polynomials form equations related to Chebyshev2 equations, with an adjustment for a power of 2.
The case then corresponds to the shortchord of a polygon, that is, the base formed by two sides of a polygon.
The comment by Peter about Lucas series, is correct, since this corresponds to a=3.  The original post corresponds to a=4.  The case where a=4, is also used to solve for Messerine primes, since where $2^p-1$ is prime, then 4^^(2^p-1) leaves a remainder of 2.
The condition amounts to this:  If the quadratic residue of (p, a^2-4) is +1, then the remainders are cyclic over a period dividing p-1.  If the residue is -1, then the period divides p+1.  If the quadratic residue of (p, a+2) is positive, the period divides the maximum period an even number of times.  If it is negative, then it divides an odd number of times.
In the case here, we have a=4, so 16-4 gives 12.  The residue against 12, is that primes (1,11) mod 12 are even divides p-1, while (5,7) are odd divides p+1.  The case of a+2=6, means that against mod 24, that primes 1,5,19,23 are even residue, so divide the maximum an even number of times, while 7,11,13,17 are odd, and divide an odd number of times.
The Chebyshev1 equations on the wiki page are offset by 1, the proper orientation is C1(0)=0, C1(1)=1, C1(2)=a, ...  , leads to the chords of a polygon, whose shortchord is 0.  When a=2, then C1(n)=n.
In this case, the chords in this format correspond to repunits of the type (1, 101, 10101, ...)  in that $C_m \mid C_n$ iff $m \mid n$, by virtue of inscribed polygons.
