As pointed out at
https://www.cut-the-knot.org/do_you_know/triSquare.shtml, square triangular numbers can be obtained by squaring the members of the sequence generated by the recurrence relation

(i) u(n) = 6*u(n-1) - u(n-2), n > 1, u(0)=0, u(1)=1

Therefore the statement can be rephrased by saying that

(ii) iff 2^p-1 is a Mersenne prime then u(2^(p-1)) = 1 (mod 2^p-1).

What is important, I think, is that u(n) is a Fibonacci type sequence, and Fibonacci type sequences reduced modulo an integer are cyclic.

For example, if p=3, then 2^p-1=7, a Mersenne prime, and u(n) modulo 7 will be:

u(0)=0 (mod 7)
u(1)=1 (mod 7)
u(2)=6 (mod 7)
u(3)=0 (mod 7)
u(4)=1 (mod 7)
u(5)=6 (mod 7), etc.

The cycle length is 3, and u(2^(p-1))=u(4) = u(1) (mod 7).
In general, if the cycle length of u(n) reduced modulo q a prime is c, then

u(1+c)=u(1) mod q, that is u(1+c)=1 (mod q).

Therefore, if I am right, what we need to prove is that

(iii) the cycle length divides 2^(p-1)-1 .

Why? With c denoting the cycle length, if c divides 2^(p-1)-1, then, for an integer k, ck+1=2^(p-1), that is, as stated in (ii), u(2^(p-1)) = u(ck+1) = 1 (mod 2^p-1).

There is an article at https://www.damtp.cam.ac.uk/user/dv211/mathgaz06.pdf, written by Dominic Vella and Alfred Vella, discussing the cycle lengths of generalised Fibonacci sequences of the type

G(n) = a*G(n-1) + b*G(n-2).

It defines Delta as a^2+4b, and then states in Theorem 6 (I only changed the letter denoting the prime):

"If Delta is a quadratic residue modulo q then C(q) | (q-1)", where C(q) is the cycle length (modulo q).

For u(n), a=6, b=-1, therefore Delta=32.
32 is a quadratic residue modulo a Mersenne prime, because there is an integer x for which
x^2 = 32 (mod 2^p-1) where 2^p-1 is a Mersenne prime. This can be rephrased as saying that there is an integer k for which k*(2^p-1)+32 is square. Take k=32=2^5, then 32*(2^p-1)+32 = 2^(p+5). As 2^p-1 is a Mersenne prime, p is odd for p>2, and so 2^(p+5) is a square for p>2.

Therefore the cycle length of u(n) modulo a Mersenne prime 2^p-1 divides 2^p-2 = 2*(2^(p-1)-1). That proves (iii). That, I think, also proves the if part of the conjecture. I am not sure of the "only if" part.

I would be grateful for a confirmation whether this really proves the if part of the conjecture.

Andras Erszegi

square triangular numbers:

https://www.cut-the-knot.org/do_you_know/triSquare.shtml

the Vella article on cycle lengths of Fibonacci sequences reduced modulo a prime:

https://www.damtp.cam.ac.uk/user/dv211/mathgaz06.pdf

I concluded that the cycle length c divides 2*(2^(p-1)-1). Then I made the error of further concluding that c divides 2^(p-1)-1. However, this is obviously false when c = 2*(2^(p-1)-1).

For instance, if we consider the following sequence:

(iv) G(n)=4*G(n-1)+4*G(n-2), n>1, G(0)=0, G(1)=1,

then theorem 6 cited from https://www.damtp.cam.ac.uk/user/dv211/mathgaz06.pdf correctly predicts that the cycle length of G(n) in (iv) modulo 31 divides 30, because Delta = 4^4+4*4 = 32 is a quadratic residue modulo 31. However, the cycle length is 30, and so it divides 2*(2^(p-1)-1), but not 2^(p-1)-1.

Therefore I did not prove the conjecture. On the other hand, I think that the approach points in the right direction.

Regards