1.3: Problem 14: Evaluate \(ab, p^4\) and \(a+b, p^4\) given that \((a,p^2) = p\) and \((b, p^3) = p^2\) where \(p\) is prime.

Solution

We are given that \((a,p^2) = p\) so \(a = pk\) for some \(k \in \mathbb{Z}\). But since \(p\) is the greatest common divisor, this also implies that \(p \not\mid k\) (since otherwise, the greatest common divisor would be \(p^2\)). Similarly, we are given that \((b,p^3)=p^2\) so \(b = p^2m\) for some \(m \in \mathbb{Z}\). Since the greatest common divisor is \(p^2\), then \(p \not\mid m\) (since otherwise, the greatest common divisor would be \(p^3\)). Therefore,

$$ \begin{align*} (ab, p^4) = ((pk) \cdot (p^2m), p^4) = (p^3 \cdot km, p^4) \end{align*} $$

Now, we know that \(p \not\mid k\) and \(p \not\mid m\) and \(p\) is prime. Therefore, \(p \not\mid km\) (By the contrapositive of the statement (if \(p\) is prime and \(p \mid ab\), then \(p \mid a\) or \(p \mid b\))). Thus, \((ab, p^4) = p^3\).

So now consider \((a + b, p^4)\). Observe that

$$ \begin{align*} (a + b, p^4) = ((pk) + (p^2m), p^4) = (p \cdot (k + pm), p^4) \end{align*} $$

We know \(p \not\mid k\). So \(p \not\mid (k+pm)\). Thus, \((a+b,p^4) = p\)

References