Find all integers n such that $2n-1 \mid n^3-3n^2+4$.

Question

Problem:

My approach:

I tried to divide $n^3-3n^2+4$ by $2n-1$ but this is not divisible. I can't get reminder as a number. I tried to manipulate the terms. I tried to factor $n^3-3n^2+4$ but failed.

Note that, since $2n-1$ is odd, i.e., has no factors of $2$, we have that $2n-1\mid n^3-3n^2+4\iff 2n-1\mid 8(n^3-3n^2+4)$. — John Omielan, Aug 23 '24 at 16:32
I tried it and it gave $2n-1 \mid 59$ but there are many numbers that can satisfy that. — , Aug 23 '24 at 16:48
It will give at most $4$ solutions but there are more.(non-positive included) — , Aug 23 '24 at 16:49
@Math12 I got a different integer than $59$, which I confirmed by rechecking, for what $2n-1$ divides. Please check your calculations to see if you perhaps made a mistake. — John Omielan, Aug 23 '24 at 16:52
As here in the dupe, by Euclid and the fractional polynomial remainder theorem: $(2n!-!1,2)!=!1\Rightarrow 2n!-!1\mid f\iff 2n!-!1\mid 2^3 f = (\color{#0a0}{2n})^3-6(\color{#0a0}{2n})^2+32\equiv \color{#c00}1^3-6(\color{#c00}1)+32\equiv27\pmod {\color{#0a0}{2n}!-!\color{#c00}1}$ — Bill Dubuque, Aug 23 '24 at 17:23
IMHO better duplicate targets include 1 and 2 though I would prefer to have a cubic divided by a linear before I use my dupehammer. — Jyrki Lahtonen, Aug 23 '24 at 17:42
This is the best match so far, I think. Adding it to the list. — Jyrki Lahtonen, Aug 23 '24 at 17:46
@Jyrki Not a good choice, since it shows only ad hoc methods, not the general way, i.e. scale so that (a fraction-free version of the) poly remainder theorem applies, as here in the 1st deup. — Bill Dubuque, Aug 23 '24 at 18:11
@Bill Tastes are different. I find that conceptually easier than, for example, introducing a non-monic variant of the division algorithm. — Jyrki Lahtonen, Aug 23 '24 at 19:20
@Jyrki It's not clear precisely what (ad hoc) method "that" refers to. The method I linked does not explicitly use nonmonic division. Rather, it simply scales so that we can apply the Polynomial Remainder Theorem (the key idea for all such problems). This is easily grasped at high-school level. — Bill Dubuque, Aug 23 '24 at 21:25
@Bill While you and I saw several concepts/rules/methodsideas//whatnot for handling elementary number theory at high school level, I find that old fashioned. For the purposes of presenting the material on this site it is more economic to only use ring/grop/polynomial ring versions of the concepts rather than referring to separate polynomial congruence rule, (fractional) polynomial CRT, etc. Unnecessary term dropping, I would say. Unnecseaary extra mental effort for a reader to try and recall the name of a simple technique rather than just apply it. — Jyrki Lahtonen, Aug 24 '24 at 02:59
You see, kids in these parts don't really see congruences and such before their freshman course. I think divisibility is touched upon in high school (casting out nines and such), but remainders are only glossed over, if that. So from my perspective learners hear about congruences on the same course they hear about groups. And the rings come up next semester. Most of the techniques you want to name are simply lumped together into "basic techniques", not worth memorizing individually, when the goal is to absorb the whole package. May be you learned these from a book that named the variants? — Jyrki Lahtonen, Aug 24 '24 at 03:08
@Jyrki That doesn't answer my question about what you mean by "that". As for your pedagogical preferences, be aware that many (most) ENT textbooks don't cover rings or groups but instead use congruence arithmetic. Congruence vs. rings first in ENT is analogous to the classical groups vs. rings first dichotomy in abstract algebra. We have to serve both schools here. I can't make any sense of your objection to naming basic results. Of course many ENT topics become unified when one studies abstract algebra. But many ENT students have not yet. We must serve them too. — Bill Dubuque, Aug 24 '24 at 03:46

J. W. Tanner · Answer 1 · 2024-08-23T17:11:13.880

-2

According to the hint from John Omielan's comment,

if $2n-1$ divides $n^3-3n^2+4$, then $2n-1$ divides $8(n^3-3n^2+4)$.

Using polynomial division, $8n^2-24n^2+32=(2n-1)(4n^2-10n-5)+27$.

Thus $2n-1$ divides $27$.

The only integers that divide $27$ are $\pm1, \pm3, \pm9, $ and $\pm27$. Can you take it from here?

edited Aug 23 '24 at 17:11

answered Aug 23 '24 at 17:09

J. W. Tanner

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Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Aug 23 '24 at 17:10
approach0 didn't find a MSE duplicate – J. W. Tanner Aug 23 '24 at 17:20
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I pointed out that it was a dupe in a comment a few minutes before your answered. If you revert to your old ways of answering many obvious dupes then you leave me no choice to to report EOQS violates, Please respect policy. – Bill Dubuque Aug 23 '24 at 17:25
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In J.W. Tanner's defence AOPS, indeed, has difficulties. Perusing the list of hits it seems to me that it cannot tell the difference between \mid and absolute value vertical separators. Therefore the list of candidates is populated by something else. The real matches may appear later. I'm not happy with an example of a gcd of a linear and a quadratic.as a duplicate target. – Jyrki Lahtonen Aug 23 '24 at 17:34
A@Jyrki Experiences users should know this is a dupe, and should know to check comments, and should be patient while others search for dupes. Not to mention it is a PSQ too. – Bill Dubuque Aug 23 '24 at 17:57
It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Aug 23 '24 at 17:57
I do agree with the sentiment that experienced user should recognize this as a dupe. However, whether J.W. Tanner is experienced enough is, in my opinion, a borderline case. Not voting on this post. – Jyrki Lahtonen Aug 24 '24 at 02:47
Should read "Approach0" instead of "AOPS" a few comments up :-) – Jyrki Lahtonen Aug 24 '24 at 02:48

Find all integers n such that $2n-1 \mid n^3-3n^2+4$.

1 Answers1