Evaluate $\gcd(ab,p^4)$ and $\gcd(a+b,p^4)$, given that $\gcd(a,p^2)=p$ and $\gcd(b,p^3)=p^2$, where $p$ is a prime.

Question

Evaluate $\gcd(ab,p^4)$ and $\gcd(a+b,p^4)$, given that $\gcd(a,p^2)=p$ and $\gcd(b,p^3)=p^2$, where $p$ is a prime. Is it true if $\gcd(a,b)=\gcd(a,c)$, then $\gcd(a^2,b^2)=\gcd(a^2,c^2)$?
We have $\gcd(a,p^2)=p\implies au_1+p^2v_1=p, u_1,v_1\in \mathbb Z\implies au_1p^2+p^4v_1=p^3\implies \gcd(a,p^4)=p^3 $

$\gcd(b,p^3)=p^2\implies bu_2+p^3v_2=p^2, u_2,v_2\in \mathbb Z\implies bu_2p+p^4v_2=p^3\implies \gcd(b,p^4)=p^3$
So $\gcd(ab, p^4)=p^3$. How to show next part?

Answer 1

By assumption, we have that there exist $x,y,z,w\in\mathbb{Z}$ so that $$ ax+p^2y=p\tag{1} $$ and $$ bz+p^3w=p^2\tag{2} $$ Separating $ax$ and $bz$ and multiplying, then moving the multiple of $p^4$ back to the left, we get $$ abxz+p^4(w+y-pwy)=p^3\tag{3} $$ Thus, $\gcd(ab,p^4)\mid p^3$. However, since $p\mid a$ and $p^2\mid b$, $p^3\mid ab$. Therefore, $$ \boxed{\displaystyle\bbox[5px]{\gcd(ab,p^4)=p^3}}\tag{4} $$

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Since $p^2\mid b$, we have $b=p^2u$, thus $(1)$ implies $$ \begin{align} \hspace{-1cm}1 &\hspace{-.7cm}=\left(\frac{a+b}px+p(y-u)\right)^3\\ &\hspace{-.7cm}=\frac{a+b}p\left[\left(\frac{a+b}p\right)^2x^3+3x^2(a+b)(y-u)+3xp^2(y-u)^2\right]+p^3(y-u)^3\\ \hspace{-1cm}p &\hspace{-.7cm}=(a+b)\left[\left(\frac{a+b}p\right)^2x^3+3x^2(a+b)(y-u)+3xp^2(y-u)^2\right]+p^4(y-u)^3\tag{5} \end{align} $$ Thus, $\gcd(a+b,p^4)\mid p$, but since $p\mid a+b$, we have $$ \boxed{\displaystyle\bbox[5px]{\gcd(a+b,p^4)=p}}\tag{6} $$

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Suppose that $\gcd(a,b)=\gcd(a,c)=d$. Then, $$ \gcd\left(\frac ad,\frac bd\right)=\gcd\left(\frac ad,\frac cd\right)=1\tag{7} $$ If $\gcd(a,b)=1$, then we have $x,y$ so that $ax+by=1$ and so $$ \begin{align} 1 &=(ax+by)^3\\ &=a^2(ax^3+3bxy)+b^2(3axy^2+by^3)\tag{8} \end{align} $$ so $\gcd(a^2,b^2)=1$. Apply this to $(7)$ and multiply by $d^2$ to get $$ \boxed{\displaystyle\bbox[5px]{\gcd(a^2,b^2)=\gcd(a^2,c^2)=d^2}}\tag{9} $$

Answer 2

Using Bezout by force seems to only complicate things here.

That $\gcd(a,p^2)=p$ means that $a=up$ where $p$ does not divide $u$. That $\gcd(b,p^3)=p^2$ means that $b=vp^2$ where $p$ does not divide $v$. Hence $ab=uvp^3$ and $p$ does not divide $uv$ since, $p$ being prime, $p\mid uv$ would imply $p\mid u$ or $p\mid v$. Thus $\gcd(ab,p^n)=p^3$ for every $n\geqslant3$.

Likewise, under the same hypotheses, $a+b=(u+vp)p$ where $u+vp$ does not divide $p$ hence $\gcd(a+b,p^n)=p$ for every $n\geqslant1$.

The second question is rather different (why asking them together?) and is probably best solved by noting that $\gcd(a^2,b^2)=\gcd(a,b)^2$ for every $(a,b)$.

Answer 3

You've got Bezout's identity wrong. $\exists u,v$ such that $au + bv = c $ only implies $\gcd(a,b) | c$, not $\gcd(a,b) = c$. I haven't found Bezout's ideneity very useful when the $\gcd$ is not 1.

$\gcd(a,p^2) = p$ means that $p$ divides $a$ and $p^2$ does not. Likewise, $p^2$ divides $b$ and $p^3$ does not. There's no way $p^3$ can divide $a$ if $p^2$ can't.

So $p^3$ divides $ab$ and $p^4$ doesn't. So $\gcd(ab,p^4) = p^3$.

We can rewrite $a = a'p$ and $b = b'p^2$, where $a'$ and $b'$ are not divisible by $p$. So $a + b = a'p + b' p^2 = p(a' + b'p)$. Then $p$ cannot divide $a' + b' p$ since p can't divide a'. So $p$ divides $a+b$ but $p^2$ can't. So $\gcd(a+b,p^3) = p$.