I have an elementary doubt, Sorry for disturbing you all. I have a statement of this sort. $r^2-1=p^a(f(p))=(r+1)(r-1). \tag{1}$ Where $r$ is an even number, and $p$ is an odd prime. $f(p)$ is a degree $n$ polynomial in $p$ with integer coefficients ( Courtesy : André Nicolas ) , $a \in \mathbb{Z}$ is a variable $>0$.
Now from the statement $(1)$ can we infer that $p^a \mid (r+1)(r-1)$ ?.
If so, $p$ being a prime, can't divide both of them (from Euclid's theorem ) . So $p^a$ divides either $(r+1)$ or $(r-1)$.
So then can we conclude that $r=kp^a \pm 1 $ for some $k>0 $?.
I told the same thing with some professor, and he is saying that it is false, and we can't write $[\ r^2-1=p^a(f(p))=(r+1)(r-1) \ ]\implies r=kp^a \pm 1.$ He gave me a counter example , which is $8\cdot6=48=2^4\cdot3$. He is now saying that my statement is like saying either $8 \text{ or } 6 $ divides $3$ which is utterly false.
So my learned elders in this community, please judge who is correct , me or the professor ?.
In any of the case, please do provide some good explanation.
Thank you.