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#### karush

##### Well-known member

- Jan 31, 2012

- 3,066

$\tiny{412.00.1.12}$

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 sorta the idea

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 sorta the idea

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