Suppose $x = p_1p_2p_3..{p_f}^n...$, so $x$ is formed of a product of some $n$ primes ($p_1,p_2$...). (The exponent of one prime is $n$; others have the exponent of 1.) Primes must start from 2, and every prime number between 2 and the last prime factor of $x$ must be a prime factor of $x$.)
Let's say that $$\beta = \frac{x}{t}$$
where $t \neq x$ and $t \neq 1$and $\beta$ must be natural number. (So $t$ must be composed of some prime factors of $x$.)
Also, let us set $\alpha = tq$. $q$ also must be a natural number.
I want to make sure that $\frac{\alpha + \beta}{2}$ would be multiples of $x$. Also, I want to make sure that $\frac{\alpha-\beta}{2}$, when factorized into prime factorization form, does not contain ${p_f}^n$.
The question is, is there any easy way to find numbers that satisfy the aforementioned for every possible $n$ (2 to infinite)?
I personally tried tackling this by setting $\alpha$ and $\beta$ as $\alpha = nx, \beta = mx$ where $n,m$ can be fractions. Would this help make cases easy?
Also, would setting $p_f$ some particular prime number help solve this case? ($p_f > 2 $)
