$\displaystyle{\displaylines{\Longrightarrow)}}$ :
Assume that :
$\displaystyle{\displaylines{\gcd(a, b) = \gcd(a, c) = 1}}$According to Bézout's theorem we have :
$\displaystyle{\displaylines{\left\{\begin{matrix}\exists (\alpha, \beta) \in \mathbb{Z}^2 \quad a \alpha + b \beta = 1\\ \exists (x, y) \in \mathbb{Z}^2 \quad a x + c y = 1\end{matrix}\right.}}$Then :
$\displaystyle{\displaylines{(\alpha a + \beta b) (xa + yc) = 1}}$Then :
$\displaystyle{\displaylines{(\alpha x a + \alpha y c + \beta b x) a + \beta bc = 1}}$We put
$\displaystyle{\displaylines{u = \alpha x a + \alpha y c + \beta b x}}$ and
$\displaystyle{\displaylines{v = \beta y}}$Then
$\displaystyle{\displaylines{\exists (u, v) \in \mathbb{Z}^2 \ : \ ua + vbc = 1}}$ According to Bézout's theorem :
$\displaystyle{\displaylines{\gcd(a, bc) = 1}}$$\displaystyle{\displaylines{\Longleftarrow)}}$ :
Assume that
$\displaystyle{\displaylines{\gcd(a, bc) = 1}}$According to Bézout's theorem :
$\displaystyle{\displaylines{\exists (u, v) \in \mathbb{Z}^2 \ : \ ua + vbc = 1}}$ we have :
$\displaystyle{\displaylines{ua + (vb)c = 1 \implies \gcd(a, c) = 1}}$we have :
$\displaystyle{\displaylines{ua + (vc)b = 1 \implies \gcd(a, b) = 1}}$Conclusion :$\displaystyle{\displaylines{\left\{\begin{matrix}\gcd(a, b) = 1 \\ \gcd(a, c) = 1\end{matrix}\right. \iff \gcd(a, bc) = 1}}$
$\displaystyle{\displaylines{\Longrightarrow)}}$ :
Assume that
$\displaystyle{\displaylines{\gcd(a, b) = 1}}$According to the first question we have :
$\displaystyle{\displaylines{\gcd(a, b^2) = 1}}$Always according to the first question:
$\displaystyle{\displaylines{\gcd(a, b^3) = 1}}$And so on until we get to :
$\displaystyle{\displaylines{\gcd(a, b^n) = 1}}$Likewise we show that :
$\displaystyle{\displaylines{\gcd(a^n, b^n) = 1}}$$\displaystyle{\displaylines{\Longleftarrow)}}$ :
Now we assume that :
$\displaystyle{\displaylines{\gcd(a^n, c^n) = 1}}$According to Bézout's theorem :
$\displaystyle{\displaylines{\exists (u, v) \in \mathbb{Z}^2 \quad u a^n + v b^n = 1}}$Then :
$\displaystyle{\displaylines{\exists (u, v) \in \mathbb{Z}^2 \quad (u a^{n-1}) a + (v b^{n-1}) b = 1}}$According to Bézout's theorem :
$\displaystyle{\displaylines{\gcd(a, b) = 1}}$
We put :
$\displaystyle{\displaylines{\gcd(a, b) = d}}$We have :
$\displaystyle{\displaylines{\gcd(a, b) = d \iff \left\{\begin{matrix}\exists (\alpha, \beta) \in \mathbb{Z}^2 \ : \ a = \alpha d, \quad b = \beta d \\ \gcd(\alpha, \beta) = 1\end{matrix}\right.}}$
Then
$\displaystyle{\displaylines{\begin{array}{rcl}\gcd(a^n, c^n) & = & \gcd((\alpha d)^n, (\beta d)^n)\\ & = & |d|^n \gcd(\alpha^n, \beta^n) \\ & = & d^n \gcd(\alpha^n , \beta^n) \\\end{array}}}$
Note that
$\displaystyle{\displaylines{\gcd(\alpha, \beta) = 1}}$, then according to the previous question :
$\displaystyle{\displaylines{\gcd(\alpha^n, \beta^n) = 1}}$Then :
$\displaystyle{\displaylines{\gcd(a^n, c^n) = d^n = \gcd(a, b)^n}}$
$\displaystyle{\displaylines{\Longrightarrow)}}$ :
Assume that
$\displaystyle{\displaylines{a | c}}$ and
$\displaystyle{\displaylines{b | c}}$ and
$\displaystyle{\displaylines{\gcd(a, b) = 1}}$We have
$\displaystyle{\displaylines{a | c}}$ then exists
$\displaystyle{\displaylines{k}}$ in
$\displaystyle{\displaylines{\mathbb{Z}}}$ that :
$\displaystyle{\displaylines{c = a k}}$Since
$\displaystyle{\displaylines{b | c}}$ we have
$\displaystyle{\displaylines{b | a k}}$We have :
$\displaystyle{\displaylines{\gcd(a, b) = 1}}$ So according to Gauss's theorem
$\displaystyle{\displaylines{b | k}}$Then exists
$\displaystyle{\displaylines{k_1}}$ in
$\displaystyle{\displaylines{\mathbb{Z}}}$ that
$\displaystyle{\displaylines{k = b k_1}}$We substitute, we have :
$\displaystyle{\displaylines{c = a b k_1}}$Then
$\displaystyle{\displaylines{ab | c}}$
$\displaystyle{\displaylines{\Longrightarrow)}}$ :
Assume that :
$\displaystyle{\displaylines{\gcd(a, b) = |a - b|}}$$\displaystyle{\displaylines{\gcd(a, b) = |a-b| \iff \left\{\begin{matrix}\exists (\alpha, \beta) \in \mathbb{Z}^{2} \ : \ a = \alpha |a-b|, \quad b = \beta |a-b| \\ \gcd(\alpha, \beta) = 1\end{matrix}\right.}}$
$\displaystyle{\displaylines{\begin{align*}\begin{cases}a=\alpha|a-b| \\ b=\beta|a-b|\end{cases} & \implies a-b=\alpha|a-b|-\beta|a-b| \\& \implies a-b=(\alpha-\beta)|a-b| \\ & \implies \alpha-\beta=\begin{cases} +1\ \text{ if : } \ a>b \\ -1 \ \text{ if : } \ a<b \end{cases} \\ & \implies \begin{cases}\beta= \alpha-1\ \text{ if : } \ a>b \\ \beta=\alpha+1\ \text{ if : } \ a<b \end{cases}\end{align*}}}$
We put:
$\displaystyle{\displaylines{\begin{cases}p=a-b \ \text{ and } \ q=+\alpha\quad \text{if : } \ a>b \\p=a-b \ \text{ and } \ q=-\alpha \quad \text{if : } \ a<b \end{cases}}}$
Then:
$\displaystyle{\displaylines{\text{if } \ a>b: \begin{cases}a=\alpha(a-b)=qp \\ b=\beta(a-b)=(q-1)p\end{cases}}}$
$\displaystyle{\displaylines{\text{if } \ a<b: \begin{cases}a=-\alpha(a-b)=qp \\ b=-\beta(a-b)=(q-1)p\end{cases}}}$
Then:
$\displaystyle{\displaylines{\gcd(a, b)=|a-b|\implies \exists p,q\in\mathbb{Z}^{*}: \begin{cases}a=pq \\ b=p(q-1) \end{cases}}}$
$\displaystyle{\displaylines{\Longleftarrow)}}$ :
Assume that:
$\displaystyle{\displaylines{\exists p,q\in\mathbb{Z}^{*}: \begin{cases}a=pq \\ b=p(q-1) \end{cases}}}$
Note that
$\displaystyle{\displaylines{\gcd(q, q-1)=1}}$ Because they are two consecutive numbers. We have :
$\displaystyle{\displaylines{\begin{align*} \gcd(q, q-1)=1 & \implies \gcd(pq, p(q-1))=|p| \\ & \implies \gcd(a, b)=|p| \end{align*}}}$
And we have
$\displaystyle{\displaylines{|a-b| = |pq-p(q-1)| = |p|}}$Then:
$\displaystyle{\displaylines{\begin{cases}a=pq \\ b=p(q-1) \end{cases} \; , \; (p,q)\in(\mathbb{Z^*})^2 \implies \gcd(a, b)=|a-b|}}$
$\displaystyle{\displaylines{\boxed{\gcd(a, b)=|a-b| \iff \exists (p,q)\in(\mathbb{Z^*})^2:\begin{cases}a=pq \\ b=p(q-1) \end{cases} }}}$
$\displaystyle{\displaylines{\Longrightarrow)}}$ :
We assume that :
$\displaystyle{\displaylines{ac \equiv bc[n]}}$, We have :
$\displaystyle{\displaylines{d=\gcd(c, n) \iff \left\{\begin{matrix}\exists (\alpha, \beta) \in \mathbb{Z}^{2} \ : \ c = \alpha d, \quad n = \beta d \\ \gcd(\alpha, \beta) = 1\end{matrix}\right.}}$$\displaystyle{\displaylines{ac\equiv bc[n]\iff \exists k\in\mathbb{Z}:(a-b)c=kn}}$$\displaystyle{\displaylines{\begin{array}{rcl}(a-b)c=kn & \iff & (a-b) \alpha d=k \beta d \\ & \iff & (a-b)\alpha=k\beta\end{array} \tag{1}}}$We have
$\displaystyle{\displaylines{\alpha|k\beta}}$ and
$\displaystyle{\displaylines{\gcd(\alpha, \beta)=1}}$ Then using Gauss's theorem we have :
$\displaystyle{\displaylines{\alpha|k}}$ ، then :
$\displaystyle{\displaylines{\exists k_1\in\mathbb{Z} \ : \ k=k_1 \alpha}}$From
$\displaystyle{\displaylines{(1)}}$ we have :
$\displaystyle{\displaylines{(a-b) = k_1 \beta}}$, then:
$\displaystyle{\displaylines{a \equiv b \left[ \beta \right]}}$Then :
$\displaystyle{\displaylines{a \equiv b \left[ \dfrac{n}{\gcd(c, n)}\right]}}$$\displaystyle{\displaylines{\Longleftarrow)}}$ :
We put :
$\displaystyle{\displaylines{d = \gcd(c, n)}}$$\displaystyle{\displaylines{\begin{array}{rcl}a\equiv b\left[\dfrac{n}{d}\right] & \implies & \exists k\in\mathbb{Z} \ : \ a-b = k\dfrac{n}{d} \\ & \implies & (a-b)c=k\dfrac{c}{d}n \\ & \implies & (a-b)c = (kc_1)n , \quad (c = c_1 \times d) \\ & \implies & ac \equiv bc \left[n\right]\end{array}}}$Conclusion:$\displaystyle{\displaylines{\boxed{ac\equiv bc[n] \iff a\equiv b\left[\frac{n}{\gcd(c, n)}\right]}}}$