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Let $n \in \mathbb{N}$. Find all $n$ satisfying that $(2n+5) \mid (5n-1)$.

Attempt: Since $2n+5$ divides $5n-1$, it must be that $$(2n+5) \mid (a(2n+5)+b(5n-1)) \ \text{for all} \ a, b \in \mathbb{N}.$$ In particular, $(2n+5) \mid ((5)(2n+5)+(-2)(5n-1))$. That is, $(2n+5) \mid 27.$

This means $2n+5 \in \{1, -1, 3, -3, 9, -9, 27, -27\}$ giving that $$n \in \{-16, -7, -4, -3, -2, -1, 2, 11\}.$$

The remaining question is : are there any other such $n$ ? Is the list above exclusive ?

I feel like what I use is just a necessary condition, $$(2n+5) \mid (5n-1) \rightarrow (a(2n+5)+b(5n-1)) \ \text{for all} \ a, b \in \mathbb{N}.$$ So I need to show that "sufficient side" as well.

Here I have that $2n+5 \mid 27$, which is just one possibility among infinitely many $a, b \in \mathbb{N}$. Somehow, if my thought is logically correct, I have that $$2n+5 \mid 27 \ \text{AND} \ (2n+5) \mid (a(2n+5)+b(5n-1)) \ \text{for all} \ a, b \in \mathbb{N}.$$ Having "AND" as the logical conjection means if the number $27$ is somehow the strictest condition among all other numbers arising form the $\mathbb{Z}-$linear combination of $2n+5$ and $5n-1$, it should imply that the list includes all possible numbers. Though I still contemplate what should be "the strictest" means, am I doing the right way ?

Bill Dubuque
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user117375
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  • You have proved that for any $n\in\mathbb{N}$, if $2n+5\mid 5n-1$, then $(2n+5)\mid 27$ and $n\in {\dots}$. This means that your list is more than exhaustive. Your only need to check which $n$ in your list actually satisfy the original constraint. – Paprika7191 Jan 12 '25 at 14:56
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    LaTeX remark: The correct command for divisibility is \mid (among others), not |. This gives correct spacing. I have replaced it. – Martin Brandenburg Jan 12 '25 at 14:58
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    (Not posting a solution as this is likely a duplicate, and definitely an abstract duplicate of GCD implications) To complete your solution along these ideas, you can further show that if $ 2n + 5 \mid 27$, then $ 2n+ 5 \mid 27 - 5(2n+5) = (-2) ( 5n-1).$ Since $2n+5$ is always odd, hence $ 2n+5 \mid 5n-1$. $\quad$ How can we generalize this approach to $ an + b \mid cb + d$? – Calvin Lin Jan 12 '25 at 18:54
  • No need to check the solutions since your first implication is an equivalence since by Euclid's Lemma $,d\mid k\iff d\mid ad!+!bk,$ when $,(d,b)=1,$ (in OP $,(d,b)=(2n!+!5,-2)=(5,-2)=1).,$ This is a special case of the (fractional) Polynomial Remainder Theorem in the linked dupe, i.e. put $,a,b = 5,-1,$ in $$,2n!+!5\mid f(n)!=!an!+!b\iff 2n!+!5,\mid, 2f\left(\small\frac{-5}2\right)!=!2\left(a\left[{\small \frac{-5}2}\right]!+!b\right) = -5a!+!2b\qquad$$ – Bill Dubuque Jan 12 '25 at 21:17
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    I don't think that this is a duplicate of the linked question, it is similar and can be solved in the same way, but not more. The mere existence of your comment (which should be an answer) that tries to explain how it follows confirms that. @BillDubuque – Martin Brandenburg Jan 12 '25 at 21:32
  • @Martin It is a special case of the Polynomial Remainder theorem in the linked dupe. We already have many tens (if not hundreds) of examples of this. There is no need for more. As for most abstract dupes, as a courtesy, I explain how it specializes in the closing comment. Please, please don't work against site organization – Bill Dubuque Jan 12 '25 at 21:43
  • Duplicate of this and this and many others. Please do not destroy site organization by reopening duplicates. $\ \ $ – Bill Dubuque Jan 14 '25 at 07:18

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You've shown (correctly, as far as I can tell) that if $2n+5 \mid (5n - 1)$, then $2n + 5 \mid 27$, so there are definitely no other values of $n$ than the ones you listed. Of course it's possible that one of the $n$ values you've found only satisfies $2n + 5 \mid 27$, but not $2n+5 \mid (5n - 1)$, so you have to check each one to be sure it's in your solution set.

This is a common practice: to find solutions to something, you find a small-ish set of potential solutions, and then use other techniques (often direct checking) to further refine the list until you're confident that all remaining items are in fact solutions.

Martin Brandenburg
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John Hughes
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  • Please strive not to post more (dupe) answers to PSQs & dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Jan 12 '25 at 21:21
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    except for this is neither a PSQ nor a duplicate. – Martin Brandenburg Jan 12 '25 at 21:33
  • @Martin But it is an (abstract) duplicate of the (fractional) Polynomial Remainder Theorem linked in the dupe - see my closing comment on the question. – Bill Dubuque Jan 12 '25 at 21:49
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    Nor does the linked answer contain my remarks about the general approach to solving "Find all x such that ..." problems by reducing the search space and then employing possibly-more-costly approaches on the remaining candidates, which attempted to at least partly address this particular OP's confusion – John Hughes Jan 13 '25 at 12:32
  • @John Not true, there are examples of that in the dupes (and that's not the question here - the OP already did that trivial part of the problem, i,e. list all divisors $,d,$ of $27$ then solve $,2n+5 = d).\ \ $ – Bill Dubuque Jan 13 '25 at 20:38
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    The confusion here was explicitly "So I need to show that sufficient side as well", i.e., the OP thinks that it's necessary for the "thin the herd" step to be a "best possible" thinning. I tried to make it clear that this is not the case (or a desirable approach in general). But you're surely right, as you always are. – John Hughes Jan 14 '25 at 04:12