Can you please help me with this question:
Show that the gcd of $8n+7$ and $7n+6$ is $1$, where $n$ is a positive integer.
Thank you.
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darkchampionz
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2Whoa...slow down a bit! You've asked (posted) five questions in the last hour. Take your time to read the answers, think about, and digest solutions and hints. If you tackle some of your earlier questions, it may help answer other questions you have. – amWhy Mar 29 '13 at 03:16
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Simply: $(8n+7)\cdot 7 - (7n+6)\cdot 8 = 1$
More generally, if $ad-bc=\pm 1$ then $\gcd(an+b,cn+d)=1$ for all integers $n$.
This is not necessary however, since $\gcd(an+0,an+1)=1$.
Thomas Andrews
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The neat thing is that if you had $an+b$ and $cn+d$ then this trick would work exactly when $ad-bc=\pm 1$ - that's the determinant of a $2\times 2$ matrix. What happens when $ad-bc\neq \pm 1$? – Thomas Andrews Mar 29 '13 at 13:36
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@vonbrand: Don't worry, I don't take offense. I essentially did the same work you did in your answer, then I hid it all in this one-line conclusion, which is definitely cheating of a sort. :) – Thomas Andrews Mar 29 '13 at 13:47
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Apply Euclid's algorithm:
$(8 n + 7) : (7 n + 6) = 1, \text{ remainder } n + 1$
$(7 n + 6) : (n + 1) = 7, \text{ remainder } -1$
So $\gcd(8 n + 7, 7 n + 6) = 1$ (it must divide $-1$ and be positive)
vonbrand
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Hint $\ $ If $\rm\:d\:$ divides both $\rm\:7+8n,\, 6+7n\:$ then eliminating $\rm\:n\:$ shows $\rm\:d\:$ divides an integer
$\begin{eqnarray} &&\rm \ &\rm\ j\,d& &\!=&\rm 7\,+\,\color{#0A0}8\,n\quad [1] \\ &&\rm &\rm k\,d& &\!=&\rm 6\,+\,\color{#C00}7\,n\quad [2]\\ \quad\color{#C00}7[1]\!-\!\color{#0A0}8[2]\:\Rightarrow\!\! &&\rm (7j\!-\!8k)\!\!\!\!\!&\rm\ \ \ d& &\!=&\rm 7(7)\!-\!8(6) = 1\:\Rightarrow\: d\mid 1 \end{eqnarray}$
Math Gems
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