I would appreciate some help finding $GCD(a^3-3ab^2, b^3-3ba^2)$; $a,b \in \mathbb{Z}$. So far I've got here: if $GCD(a,b)=d$ then $\exists \alpha, \beta$ so that $GCD(\alpha, \beta)=1$ and $\alpha d=a$, $\beta d=b$.

Therefore we know that $GCD(a^3-3ab^2, b^3-3ba^2)=d^3 GCD(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2)$. However I don't know to figure out $GCD(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2)$ given that $GCD(\alpha,\beta)=1$.

  • $\begingroup$ I meant that a and b are integers. $\endgroup$ – Oleksandr Mar 21 at 16:31
  • $\begingroup$ For integers, it can have many values. What do you want to prove? $\endgroup$ – Dietrich Burde Mar 21 at 16:38
  • $\begingroup$ I want to find the explicit formula for GCD of those two polynomials in terms of GCD of a and b. $\endgroup$ – Oleksandr Mar 21 at 16:48

As you've already shown, factoring out the cube of the GCD of $a$ and $b$ gives a new equation

$$e = \gcd\left(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2 \right) \tag{1}\label{eq1}$$


$$\gcd\left(\alpha,\beta \right) = 1 \tag{2}\label{eq2}$$

Update: Here is a simpler solution than what I originally wrote. First, note that no factor of $e$ may divide $\alpha$ or $\beta$. If any do, let's say $\alpha$, then it must divide $\beta^3 - 3\beta\alpha^2$ and, thus, must divide $\beta^3$, which is not possible due to \eqref{eq2}. Thus, from the first term of \eqref{eq1}, since $\alpha^3-3\alpha \beta^2 = \alpha\left(\alpha^2 - 3\beta^2\right)$, this means that $e \mid \alpha^2 - 3\beta^2$. Similarly, for the second term, $\beta^3-3\beta \alpha^2 = \beta\left(\beta^2 - 3\alpha^2\right)$ gives that $e \mid \beta^2 - 3\alpha^2$. Also, $e$ must divide any linear combination of these values, including $e | \alpha^2 - 3\beta^2 + 3\left(\beta^2 - 3\alpha^2\right) = -8\alpha^2$. Thus, $e$ can only be a power of $2$. To finish this off, go to the second last paragraph. Otherwise, for the rest of the original, longer solution, continue reading.

$ $

Next, note that if $f = \gcd(g,h)$, then $f$ divides $g$ and $h$ and, thus, will also divide any linear combination of $g$ and $h$, including their sum & difference. From \eqref{eq1}, first check the sum of the $2$ inside values:

\begin{align} \alpha^3-3\alpha \beta^2 + \beta^3-3\beta \alpha^2 & = \alpha^3 + \beta^3 -3\left(\alpha \beta\right)\beta - 3\left(\alpha \beta\right)\alpha \\ & = \left(\alpha + \beta\right)\left(\alpha^2 - \alpha\beta + \beta^2\right) - 3\alpha\beta\left(\alpha + \beta\right) \\ & = \left(\alpha + \beta\right)\left(\alpha^2 - 4\alpha\beta + \beta^2\right) \tag{3}\label{eq3} \end{align}

Suppose there's a factor $m \gt 1$ which divides $e$ and $\alpha + \beta$. Then, $\alpha \equiv -\beta \pmod m$, so $\alpha^3-3\alpha \beta^2 \equiv 2\beta^3 \pmod m$. From \eqref{eq2}, this means that $m = 2$, and that any other factors of $e$ must divide $\alpha^2 - 4\alpha\beta + \beta^2$.

From \eqref{eq1}, next check the difference of the $2$ inside values:

\begin{align} \alpha^3-3\alpha \beta^2 - \beta^3 + 3\beta \alpha^2 & = \alpha^3 - \beta^3 -3\left(\alpha \beta\right)\beta + 3\left(\alpha \beta\right)\alpha \\ & = \left(\alpha - \beta\right)\left(\alpha^2 + \alpha\beta + \beta^2\right) + 3\alpha\beta\left(\alpha - \beta\right) \\ & = \left(\alpha - \beta\right)\left(\alpha^2 + 4\alpha\beta + \beta^2\right) \tag{4}\label{eq4} \end{align}

Suppose there's a factor $n \gt 1$ which divides $e$ and $\alpha - \beta$. Then, $\alpha \equiv \beta \pmod n$, so $\alpha^3-3\alpha \beta^2 \equiv -2\beta^3 \pmod m$. From \eqref{eq2}, this means that $n = 2$, and that any other factors of $e$ must divide $\alpha^2 + 4\alpha\beta + \beta^2$.

This shows that any factor, other than $2$, which divides $e$ must divide both $\alpha^2 - 4\alpha\beta + \beta^2$ and $\alpha^2 + 4\alpha\beta + \beta^2$. Thus, it must also divide their difference, which is $8\alpha\beta$. This can only be true for $2$, $4$ or $8$.

At this shows overall, only powers of $2$ may possibly divide $e$. Since $e$ is relatively prime to $\alpha$ & $\beta$, this means they must both be odd. From $\alpha^3 - 3\alpha\beta^2 = \alpha\left(\alpha^2 - 3\beta^2\right)$, note that $\alpha^2 \equiv \beta^2 \equiv 1 \pmod 4$, so $\alpha^2 - 3\beta^2 \equiv -2 \pmod 4$. In other words, there will only be $1$ factor of $2$.

In summary, with your original equation of $d = \gcd\left(a,b\right)$, we get that $\gcd\left(a^3-3ab^2, b^3-3ba^2\right)$ is $2d^3$ if both $\frac{a}{d}$ and $\frac{b}{d}$ are odd, else it's $d^3$.

  • $\begingroup$ Wow, this is an outstanding explanation and it completely solves the problem. Thank you very much. $\endgroup$ – Oleksandr Mar 21 at 18:28
  • 1
    $\begingroup$ @Oleksandr You are welcome. I had put in, removed (as I temporarily thought it was wrong) and now put back a shorter, simpler solution near the start of the answer. Sorry for any confusion my back & forth may have caused with this. $\endgroup$ – John Omielan Mar 21 at 18:54

Note that $\alpha^3 - 3 \alpha \beta^2$ and $3\beta\alpha^2 - \beta^3$ are the real and imaginary parts respectively of $(\alpha + i \beta)^3$; this suggests that it will be useful to work in the ring of Gaussian integers $\mathbb{Z}[i]$.

Now, suppose that for some integer prime $p$, $p \mid \alpha^3 - 3\alpha \beta^2$ and $p\mid 3\beta\alpha^2 - \beta^3$; then $p \mid (\alpha + i \beta)^3$. If $p \equiv 3 \pmod{4}$, then $p$ remains irreducible in $\mathbb{Z}[i]$, so $p \mid \alpha + i \beta$, implying that $p \mid \gcd_{\mathbb{Z}}(\alpha, \beta)$ which gives a contradiction. Similarly, if $p \equiv 1 \pmod{4}$, then the factorization of $p$ into irreducibles is of the form $p = (c + di) (c - di)$ for some $c, d \in \mathbb{Z}$. Thus, $c + di \mid (\alpha + i \beta)^3$ implies $c + di \mid \alpha + i\beta$ and $c - di \mid (\alpha + i\beta)^3$ implies $c - di \mid \alpha + i \beta$. Since $c + di$ and $c - di$ are relatively prime in $\mathbb{Z}[i]$ (being irreducibles which do not differ by multiplication by a unit), this implies $(c + di) (c - di) \mid \alpha + i \beta$; in other words, $p \mid \alpha + i \beta$, again giving a contradiction.

The only remaining possibility is $p = 2$ which factors as $-i (1+i)^2$ in $\mathbb{Z}[i]$. Now, by a similar argument to the above we have $(1 + i)^2 \nmid \alpha + \beta i$, so the order of $1+i$ in $(\alpha + \beta i)^3$ is either 0 or 3. In the former case we get that $\gcd(\alpha^3 - 3\alpha\beta^2, 3\beta\alpha^2 - \beta^3) = 1$, and in the latter case we get that $\gcd(\alpha^3 - 3\alpha\beta^2, 3\beta\alpha^2 - \beta^3) = 2$.

Note that this solution is very closely tied to the exact form of the polynomials under consideration; whereas the general idea of John Omielan's answer should be more generally applicable to other cases.


I'll write $\alpha=m,\beta=n$ for the ease of typing

If $d(\ge1)$ divides $m^3-3mn^2,n^3-3m^2n$

$d$ will divide $n(m^3-3mn^2)+3m(n^3-3m^2n)=-8m^3n$

and $3n(m^3-3mn^2)+m(n^3-3m^2n)=-8mn^3$

Consequently, $d$ must divide $(-8m^3n,-8mn^3)=8mn(m^2,n^2)=8mn$

So, $d$ will divide $8$

As $(m,n)=1,$ both $m,n$ cannot be even

If $m$ is even, $n^3-3m^2n$ will be odd $\implies d=1$

If both $m,n$ are odd,. $m^2,n^2\equiv1\pmod8$

$m(m^2-3n^2)\equiv m(1-3)\equiv-2m\pmod8$

Similarly, we can establish the highest power of $2$ in $m^3-3mn^2$ will be $2$

$\implies d=2$ if $m,n$ are odd


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