Exercise 1.3.18 ($a\wedge b = c$ implies $a^2 \wedge b^2 = c^2$)

Question

Prove that $(a^2,b^2) = c^2$ if $(a,b) = c$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $a,b,c$ be positive integers. Write
$$a = \prod_{p \in A} p^{\alpha(p)},\qquad b = \prod_{p \in A}p^{\beta(p)}, \qquad c = \prod_{p \in A}p^{\gamma(p)}.$$
(We can take the same set $A$, if we put $\alpha(p) \geq 0, \beta(p) \geq 0, \gamma(p) \geq 0$.)

Suppose that $a \wedge b = c$. Then $\gamma(p) = \min(\alpha(p), \beta(p))$ (see (1.7)).

Hence, using (1.7) anew,
\begin{align*}
a^2 \wedge b^2 &= \prod_{p\in A} p^{\min(2\alpha(p), 2 \beta(p))}\
&= \prod_{p\in A} p^{2\min(\alpha(p),  \beta(p))}\
&=\prod_{p\in A} p^{2\gamma(p)}\
&=\left( \prod_{p\in A} p^{\gamma(p)}ight)^2\
&=c^2.
\end{align*}
So 
$$a\wedge b = c \Rightarrow a^2 \wedge b^2 = c^2.$$
\end{proof}

Exercise 1.3.18 ($a\wedge b = c$ implies $a^2 \wedge b^2 = c^2$)

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Exercise 1.3.18 ($a\wedge b = c$ implies $a^2 \wedge b^2 = c^2$)

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