I have created a hypothesis:
$$\text{If } p \text{ is a perfect number, then } 1 + 8p \text{ is a squared number.}$$
Could anyone prove/disprove the above statement? I have no idea where to begin, and I discovered this by pure accident when trying to solve a quadratic equation.
Edit:
I believe that all odd numbers $m$ squared can be expressed as the form $8n + 1$ for some $n \in \mathbb{N}$.
$$\begin{align} 3^2 &= 9 = 8\times 1 + 1 \\ 5^2 &= 25 = 8\times 3 + 1 \\ 7^2 &= 49 = 8\times 6 + 1 \end{align}$$
Yes, it is obvious that $n$ is also a triangular number $(1 + 2 + 3 + \cdots)$. Let $m = 2k + 1$, then one has the following: $$(2k + 1)^2 = 8n + 1$$ And $(2k + 1)^2 = 4k(k + 1) + 1$, so it follows, then, that:
$$4k(k + 1) \require{cancel}\cancel{+1} = 8n \cancel{+1} \Rightarrow k(k + 1) = k^2 + k = 2n$$
And since the parity of $r \in \mathbb{N}$ remains the same for $r^k$ for all $k \in \mathbb{N}$ then our proof stands valid.
This is true for even perfect numbers, at least. Every even perfect number can be written as $2^{n-1} (2^n - 1)$, for some $n$, which is necessarily a triangle number. $29^2 + 6728p = 29^2 (1 + 8p)$, but it is a well known (and easy to prove) fact that if $p$ is triangular, then $1+8p$ is a square.