I wish to show that the numbers $5a+2$ and $7a+3$ are relatively prime for all positive integer $a$.
Here are my solutions.
Solution 1. I proceed with Euclidean Algorithm. Note that, for all $a$, $|5a+2|<|7a+3|$. By Euclidean Algorithm, we can divide $7a+3$ by $5a+2$. To have
$7a+3=(5a+2)(1)+(2a+1)$ continuing we have,
Since the last nonzero remainder in the Euclidean Algorithm for $7a+3$ and $5a+2$ is 1, we conclude that they are relatively prime.
Solution 2. Suppose that $d=\gcd(5a+2,7a+3)$. Since $d=\gcd(5a+2,7a+3)$ then the following divisibility conditions follow:
(1) $d\mid (5a+2)$
(2) $d\mid (35a+14)$
(3) $d\mid (7a+3)$
(4) $d\mid (35a+15)$.
Now, (2) and (4) implies that $d$ divides consecutive integers. The only (positive) integer that posses this property is $1$. Thus, $d=1$ and that $7a+3$ and $5a+2$ are relatively prime.
Here are my questions:
Is the first proof correct or needs to be more specific? For instance cases for $a$ must be considered.
Which proof is better than the other?
Thank you so much for your help.