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I have to solve in $\mathbb{N}$ the following:

$$3n - 14m = 2$$

So basically this means $$n = \dfrac{2}{3}(7m+1)$$

which means I need $7m - 1$ do be a multiple of $3$. For example, it happens when $m = 2$ and $m = 5$ and basically for $m = 3k-1$ for $k = 1, 2, \cdots$. I have found this by a hand trying, nothing else.

Is there any way to "solve it better" or have I to get settled with this only?

Bill Dubuque
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Heidegger
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1 Answers1

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Since the greatest common divisor of $2$ and $3$ is $1$, then $$\frac{2(7m-1)}{3}\in\mathbb{Z}\Longleftrightarrow \frac{7m-1}{3}\in\mathbb{Z}\Longleftrightarrow$$$7m-1$ is a mutiple of $3$.

Then we note that for a given integer $m_0$ , $7m_0-1$ is a multiple of $3\Longleftrightarrow$ for all integer $k$, $7(3k+m_0)-1$ is a multiple of $3$, which uses the fact that $1$ is the greatest common divisor of $7$ and $3$. So we conclude that the solutions of $m$ must be the unions of several sets $\{3i+k| k$ is a fixed integer $\}$. Then it remains to check if $m=1,2,3$ works, because the set$\{3i+k_1\}$ and $\{3i+k_2\}$ are the same if $k_1-k_2$ is a multiple of $3$.

J. W. Tanner
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Asigan
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 13 '22 at 14:24
  • @BillDubuque Well, but you should note that I wrote my answer before it was labeled duplicate. – Asigan Nov 13 '22 at 14:44
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    That's no excuse - it should be clear that a question so basic has been already answered many times on a site more than 12 years old. – Bill Dubuque Nov 13 '22 at 14:51
  • @BillDubuque Fine, what about posting comments under the question? Please tell me if this is allowed. – Asigan Nov 14 '22 at 01:25