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The question was:

  • Find $n$ where $GCD(a_{n}, 14) = 7$ where $n$ is natural if you knew that $a_{n} = n + 3$

The book solved it by saying that means $a_{n}$ is a multiple of $7$ but not $14$ so $a_{n} = 7a$ and by putting it in the equation

$GCD(7a, 14) = 7$ means $GCD(a,2) = 1$ which means $a = 2k + 1$ and by putting it in $a_{n}$ we find that n = 14a + 4$

The first, I understand that because if it was a multiple of 14 then we would take $14$ as the $GCD$ not $7$

What I didn't understand is that what putting $a_{n} = 7a$ has anything to do with not being a multiple of 14? 14 can be written as $7(2)$ which means $a_{n}$ can be clearly a multiple of 14 too.

Bill Dubuque
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TechnoKnight
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  • Use \gcd for $\gcd$, or, if you really want to use capital letters, use \operatorname{GCD} for $\operatorname{GCD}$, though this is less common. – DMcMor Feb 02 '21 at 20:21
  • My guess would be that they mean $a_n=7a$ with odd $a$. But that's something they should -say-, not leave to your imagination. (Myself I'd write $a_n=14b+7$. That's the same if you write $a=2b+1$ but makes $a$ explicitly odd.) – Semiclassical Feb 02 '21 at 20:22
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    It is clearly intended that $a_n$ expressible in the form $(7a)$ is a necessary condition. This is how you take a 1st step in attacking the problem. The next step would be to refine the constraint by further specifying that $a$ is odd. The misinterpretation that you encountered was that you took the constraint $a_n = 7a$ out of context and misinterpreted it as intending that the constraint (in addition to being necessary) was also sufficient. – user2661923 Feb 02 '21 at 20:44
  • If $a_n$ is a multiple of $14$, then $GCD(a_{n}, 14) = 14$. – Steven Alexis Gregory Feb 07 '21 at 03:46

2 Answers2

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The argument you quote that $GCD(7a, 14) = 7$ means the same as $GCD(a, 2) = 1$ and therefore $a=2k+1$ is correct.

You therefore know that $$n+3=7a=7(2k+1)=14k+7.$$

Then $n=14k+4$, as required

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The proof is incorrect / incomplete since it gives only a necessary condition for $n$ to be a solution. To fix it you either need to check that the candidate $n$ works or else verify that all inferences are actually bidirectional (see here for more on the insufficiency of unidirectional inferences). Let's do the proof with bidirectional inferences, which will answer your question along the way.

First, $7$ is a common divisor of $\,n+3\,$ and $\,14,\,$ thus a divisor of $\,n+3,\,$ i.e. $\,\color{#0a0}{7\mid n+3},\,$ Using this and $\, 7x = (7y,7z)\color{#c00}\iff x = (y,z)\,$ by $\rm\color{#c00}{GDL}$ = gcd distributive law we have

$$\begin{align} 7 = (n+3,14) &\iff \ \ \color{#0a0}{7\mid n+3}\ \ \ \ \ \&\ \ \ 7 = (n+3,\ \ 14)\\[.1em] &\iff n=4+7k\ \ \ \&\ \ \ 7 = (7+7k,\:\!14)\\[.1em] &\color{#c00}\iff n=4+7k\ \ \ \&\ \ \ 1 = (1+k,\ 2)\ \ \ \rm by\ \color{#c00}{GDL}\\[.1em] &\iff n=4+7k\ \ \ \&\ \ \ k = 2j\\[.1em] &\iff n = 4+7(2j) \end{align}\qquad$$

Bill Dubuque
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