412.00.1.12 are relatively prime for all n.

[karush]

$\tiny{412.00.1.12}$
Show that $5n+3$ and $7n+4$ are relatively prime for all n.
$$ax + by = 1$$
$\begin{array}{ll}
\textit{let} &a=5n+3 \textit{ and } b=7n+4\\
\textit{then} &(5n+3)x + (7n+4)y = 1\\
\textit{compute}&(7n+4)=(5n+3)+(2n+1)\\
&(5n+3)=2\cdot(2n+1)+(n+1)\\
&(5n+3)-(4n+2)=n+1
\end{array}$

ok no book answer but hope what I put here is sort of the idea

 

[Opalg]

$7(5n+3) - 5(7n+4) = 1$.