Given $p \in \mathbb{N}$, $p > 1$ and the set: $$A_p = \{x \ | \ (\exists) \ m ,n \in 2\mathbb{N} + 1, x = (1 + 3 + \dots + m)p + (m + 2) + (m + 4) + \dots + n\}$$ Prove that, if $x, y, \in A_p$, then $xy \in A_p$.
Source: Romanian magazine in mathematics.
Attempt: Note that $1 + 3 + \dots + k = z^2$, if $k \in 2\mathbb{N} + 1$. So, the set $A_p$ may be redefined as: $$A_p = \{x \ | \ x = p\alpha^2 + \beta^2 - \alpha^2, \alpha, \beta \in \mathbb{N}^* \}$$
Therefore, $x = (p - 1)\alpha^2 + \beta^2$ and $y = (p - 1)\gamma^2 + \delta^2$. So, $xy$ should be written as $(p - 1)\varepsilon^2 + \varphi^2$, but multiplying the $x$ and $y$ equations, the general form doesn't look good.
I'm also thinking at induction, as $p$ natural. We may start with $p = 2$ (it may be simpler) and than a general $p \to p + 1$ case.
Use $(a^2+n b^2)(c^2+n d^2) = (ac+nbd)^2 + n(ad-bc)^2$