# Valid proof for (2n+1,3n+1)=1?

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

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

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

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