$x,y,z$ are integers
I think the answer is if the $\gcd(a,b,c)|d$, but I don't know how to show this in either direction
Asked
Active
Viewed 257 times
0
-
Here's a start. Let $m = \gcd(a,b,c)$. Then $m|ax+by+cz$ so, we must also have $m|d$. – videlity Sep 26 '17 at 02:59
-
1See generalizations of Bezout's lemma. The proof should be similar to the proof for the case of two integers. – Prasun Biswas Sep 26 '17 at 03:04
-
@videlity Ok that makes perfect sense since it divides one side of the equation it must divide the other side. Can I have a hint for the reverse direction? I need if and only if – Vinny Chase Sep 26 '17 at 03:05
-
@VinnyChase I believe the proof is a constructive one using the extended Euclidean algorithm/Bezout's lemma. – videlity Sep 26 '17 at 03:13
-
The condition is clearly necessary, and it is sufficient by Bezout's gcd identity, which works for any number of arguments (a simple, conceptual proof.is here.) – Bill Dubuque Sep 26 '17 at 03:28