Perfect numbers and Pell's equation

Question

(Disclaimer: The following is a naive attempt to apply the theory of Pell's equations to perfect numbers. Please bear in mind that this is my first time to try solving such an equation.)

Let $p^k$ be a nontrivial prime power satisfying $p \equiv k \equiv 1 \pmod 4$. (That is, we require $k > 1$.)

Suppose that $M = 2^{t-1}(2^t - 1)$ is an even perfect number. By the Euclid-Euler Theorem, $2^t - 1$ (and therefore $t$) must be prime. (If $t$ is prime, it does not necessarily follow that $2^t - 1$ is also prime.)

Here is my:

INITIAL QUESTION

Is $M(2p^k - 1) + p^{2k}$ a triangular number?

MY ATTEMPT

Let $$T = M(2p^k - 1) + p^{2k}.$$ Then $T$ is triangular if and only if $$1 + 8T = S^2.$$ Suppose to the contrary that $T = U^2$ is also a square. It follows that $M > 6$, because otherwise when $M = 6$, we obtain $$M(2p^k - 1) + p^{2k} \equiv 2\cdot(2\cdot{1} - 1) + 1^2 \equiv 3 \pmod 4.$$ Then we obtain the Pell's equation $$S^2 - 8U^2 = 1.$$
This has fundamental solution $(S,U)=(3,1)$. However, we also know that $$T = 2^{t-1}(2^t - 1)(2p^k - 1) + p^{2k}$$ which appears not to split into terms linear in $2^{t-1}$, $2^t - 1$ and $p^k$, for $t \geq 3$ and $k > 1$, at least per WolframAlpha.

This hints that $T$ may not be a square, hence the theory of Pell's equations may not apply. (Note that the lowest possible value of $T$ is $9803119 = {47} \times {208577}$, which is not triangular. Hence I am led to believe that, in general, $T$ may, in fact, neither be triangular nor a square.)

FINAL QUESTIONS

(1) Does my proof for showing that $T$ is not a square triangular number suffice? If the proof is not logically correct, how can it be mended so as to produce a valid argument?

(2) Would it then be possible to (unconditionally) show that $M(2p^k - 1) + p^{2k}$ is not triangular?

Answer 1

Partial answer: The following PARI/GP routine searches for examples:

gp > forprime(q=1,1000,if(isprime(2^q-1,2)==1,s=(2^q-1)*2^(q-1);forprime(p=1,10^8,if(Mod(p,4)==1,forstep(k=1,30,4,t=s*(2*p^k-1)+p^(2*k);if(issquare(t)+issquare(8*t+1)>0,print([s,p,k,t,factor(t)])))))))
[496, 5, 1, 4489, Mat([67, 2])]
[496, 1733, 1, 4721929, [41, 2; 53, 2]]
[8128, 16510001, 1, 272848519588129, [157, 2; 105211, 2]]
gp >

For all solutions in this range, we have $\ k=1\ $ (which is however ruled out). Hence this first range shows that for a solution, we must have at least one of $\ M>10^{769}\ $ , $\ p>10^8\ $ , $\ k>30\ $

gp > forprime(q=1,1000,if(isprime(2^q-1,2)==1,s=(2^q-1)*2^(q-1);forprime(p=1,10^4,if(Mod(p,4)==1,forstep(k=1,4000,4,t=s*(2*p^k-1)+p^(2*k);if(issquare(t)+issquare(8*t+1)>0,print([s,p,k,t,factor(t)])))))))
[496, 5, 1, 4489, Mat([67, 2])]
[496, 1733, 1, 4721929, [41, 2; 53, 2]]
gp >

This second search range also reveals no solutions and shows that we must have at least one of $\ M>10^{769}$ , $\ p>10^4\ $ , $\ k>4\ 000\ $

I will further examine the $\ k=5\ $-case. Since $\ p>10^8\ $ is established, we know that a solution must exceed $\ 10^{40}\ $ making it unlikely that there is one.

Answer 2

Too long to comment :

For $T$ to be a square, it is necessary that $$p^k\leqslant 2^{t-2}(2^t-1)\bigg(2^{t-1}(2^t-1)-1\bigg)-1$$

Proof :

Since $T$ can be written as $$T = \bigg(p^k+2^{t-1}(2^t - 1)\bigg)^2-2^{t-1}(2^t-1)\bigg(2^{t-1}(2^t-1)+1\bigg) $$ one can see that $$T\lt \bigg(p^k+2^{t-1}(2^t - 1)\bigg)^2$$ holds.

So, one can see that if $$\bigg(p^k+2^{t-1}(2^t - 1)-1\bigg)^2\lt T\tag1$$ holds, then since $$\bigg(p^k+2^{t-1}(2^t - 1)-1\bigg)^2\lt T\lt \bigg(p^k+2^{t-1}(2^t - 1)\bigg)^2\tag2$$ holds, $T$ cannot be a square.

Now, $(1)$ is equivalent to

$$\begin{align}&2^{t-1}(2^t-1)\bigg(2^{t-1}(2^t-1)+1\bigg)\lt 2\bigg(p^k+2^{t-1}(2^t - 1)\bigg)-1 \\\\&\iff 2p^k\gt 2^{t-1}(2^t-1)\bigg(2^{t-1}(2^t-1)-1\bigg)+1 \\\\&\iff 2p^k\geqslant 2^{t-1}(2^t-1)\bigg(2^{t-1}(2^t-1)-1\bigg)+2 \\\\&\iff p^k\geqslant 2^{t-2}(2^t-1)\bigg(2^{t-1}(2^t-1)-1\bigg)+1\tag3\end{align}$$

In short, if $(3)$ holds, then since $(2)$ holds, $T$ cannot be a square.

Therefore, for $T$ to be a square, it is necessary that $$p^k\lt 2^{t-2}(2^t-1)\bigg(2^{t-1}(2^t-1)-1\bigg)+1\tag4$$

Since the both sides of $(4)$ are odd, $(4)$ is equivalent to $$p^k\leqslant 2^{t-2}(2^t-1)\bigg(2^{t-1}(2^t-1)-1\bigg)-1\ .\quad\blacksquare$$