Prove that if $n$ is divisible by a prime number $p$ then neither $n^2 +1$ nor $n^2 -1$ will be divisible by $p$.

Question

I know this holds for $p=3$, but can it be generalized for any prime number? Can it be generalized further for any integer $p \in \Bbb N $ ?

Answer 1

If integer $p$ divides both $n$ and $(n^2-1)(n^2+1)$

$p$ must divide $$n\cdot n^3-(n^2-1)(n^2+1)=?$$

Answer 2

If $p$ divides $n$ wouldn't $p$ divide $n^2$ as well. If $p$ a prime number divides $n^2$ will it divide $n^2+1$ or $n^2-1$? The lowest prime number is $2$.

Answer 3

Hint:-

We have $n$ is divisible by $p$.

So,$n^2$ must be divisible by $p$.(why?)

So,clearly $n^2+1$ will leave remainder $1$.(why?)

Now,try to find out what remainder $n^2-1$ will leave.