If $a$ is an integer, prove that $\gcd(14a + 3, 21a + 4) = 1$.

Question

If $a$ is an integer, prove that gcd$(14a+3,21a+4)=1$.

I want to know if my proof is correct or if I should start over because I did something wrong.

My proof is as followed:

Assume $a \in Z$. Wts: gcd$(14a+3,21a+4)=1$. By theorem we know that $(14a+3)x+y(21a+4)=1$. So, let $x=3$ and $y=-2$. Thus, $(14a+3)(3)+(-2)(21a+4)=1$. Hence, gcd$(14a+3,21a+4)=1$.

Answer 1

Yes, your solution is correct. It follows from $$(14a+3)(3)+(-2)(21a+4)=1$$

There is no need for assumption of existence of $x$ and $y$. You showed that they exist, namely $x=3$ and $y=-2$.

Answer 2

You do not need the first sentence to prove the result. Start with " Let x=3 and y=-2...." and your proof will be good.

Answer 3

$\textbf{Idea:}$

$$(14a+3)(3)+(21a+4)(-2) = 1.$$

We know that $\gcd((14a + 3), (21a + 4)) = n \geq 1$, then $n|(14a+3)$ and $n|(21a+4)$. Therefore $n|1 \therefore n = 1$.

Answer 4

You can use that

$$\gcd(a,b)=\gcd(a,sa+tb)$$

then

$$\gcd(14a+3,21a+4)=\gcd(42a+9,42a+8)=\gcd(42a+9-42a-8,42a+8)=\\=\gcd(1,42a+8)=1$$