Is this a Valid proof for $(2n+1,3n+1)=1$? Is this a valid proof for $(2n+1,3n+1)=1$?

$\exists \ d \in \mathbb{Z}$


$d | 2n+1$ means $2n+1 \equiv 0$ (mod $d$)
$d | 3n+1$ means $3n+1 \equiv 0$ (mod $d$)


Multiply 1. by $3$ and 2. by $2$:

$6n+3 \equiv 0$ (mod $d$)

$6n+2 \equiv 0$ (mod $d$)

Subtract the second from the first to get:
$ 1 \equiv 0$ (mod $d$), which means $d|1$, which means $d=1$.

Thanks!
Yes, nice job! That's a very nice application of the method that I described in your prior question. Notice again how simple the problem becomes once it is reduced to equational (congruence) form.
@Bill has already commented on the truth of your proof, here is another one.

Suppose that $(2n+1,3n+1)=d$.

$$d|3n+1, d|2n=1$$

$$d|3n+1-(2n+1)\implies d|n\implies d|2n$$

$$d|2n+1-2n\implies d|1\implies d=1$$