Greatest Common Divisor Problem: Prove that $\gcd(\frac{a^3+b^3}{a+b}, a+b) = \gcd(a+b, 3ab)$

Question

I've been stuck in this problem for some time now. Currently what I have accomplished is, using the propriety $\gcd(a,b) = \gcd(b,a \bmod(b))$ to get in the equation

$$ \gcd(a+b, \frac{a^3+b^3}{a+b}\bmod(a+b)) $$

but I don't know where to go anymore or even if I'm in the correct path. Any tips or solutions would be great appreciated.

Note that $\frac{a^3+b^3}{a+b}=a^2-ab+b^2$ – Angina Seng Jul 27 '20 at 20:16 — Angina Seng, Jul 27 '20 at 20:16

score 4 · Accepted Answer · answered Jul 27 '20 at 20:17

4

hint: $a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)((a+b)^2 -3ab)$

answered Jul 27 '20 at 20:17

openspace

6,470

score 2 · Answer 2 · answered Jul 27 '20 at 23:30

2

We have $gcd((a^3 + b^3)/(a+b), a + b) = gcd(a^2 - ab + b^2, a + b) = gcd((a^2 - ab + b^2) - (a+b)(a+b), a + b) = gcd(-3ab, a + b) = gcd(3ab, a + b)$.

answered Jul 27 '20 at 23:30

Doctor Who

3,461

score 1 · Answer 3 · answered Jul 27 '20 at 21:03

We have $$ \begin{pmatrix} \frac{a^3+b^3}{a+b} \\ a+b \end{pmatrix} = \begin{pmatrix} a^2-ab+b^2 \\ a+b \end{pmatrix} = \begin{pmatrix} -1 & a+b \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 3ab \\ a+b \end{pmatrix} $$ The result follows because the matrix has determinant $-1$ and so has an integer inverse.

Greatest Common Divisor Problem: Prove that $\gcd(\frac{a^3+b^3}{a+b}, a+b) = \gcd(a+b, 3ab)$

3 Answers3