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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Inspired by 2025 Beijing
sqing   10
N an hour ago by ytChen
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
10 replies
sqing
Yesterday at 4:56 PM
ytChen
an hour ago
Three operations make any number
awesomeming327.   0
3 hours ago
Source: own
The number $3$ is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.
0 replies
1 viewing
awesomeming327.
3 hours ago
0 replies
IMO 2017 Problem 4
Amir Hossein   116
N 3 hours ago by cj13609517288
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
116 replies
Amir Hossein
Jul 19, 2017
cj13609517288
3 hours ago
A sharp one with 3 var
mihaig   10
N 3 hours ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
10 replies
mihaig
May 13, 2025
mihaig
3 hours ago
Another right angled triangle
ariopro1387   1
N 3 hours ago by lolsamo
Source: Iran Team selection test 2025 - P7
Let $ABC$ be a right angled triangle with $\angle A=90$. Point $M$ is the midpoint of side $BC$ And $P$ be an arbitrary point on $AM$. The reflection of $BP$ over $AB$ intersects lines $AC$ and $AM$ at $T$ and $Q$, respectively. The circumcircles of $BPQ$ and $ABC$ intersect again at $F$. Prove that the center of the circumcircle of $CFT$ lies on $BQ$.
1 reply
ariopro1387
Today at 4:13 PM
lolsamo
3 hours ago
four points lie on a circle
pohoatza   78
N 4 hours ago by ezpotd
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
78 replies
pohoatza
Jun 28, 2007
ezpotd
4 hours ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   2
N 4 hours ago by Stear14
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
2 replies
FishkoBiH
Today at 1:38 PM
Stear14
4 hours ago
Does there exist 2011 numbers?
cyshine   8
N 4 hours ago by TheBaiano
Source: Brazil MO, Problem 4
Do there exist $2011$ positive integers $a_1 < a_2 < \ldots < a_{2011}$ such that $\gcd(a_i,a_j) = a_j - a_i$ for any $i$, $j$ such that $1 \le i < j \le 2011$?
8 replies
cyshine
Oct 20, 2011
TheBaiano
4 hours ago
D1036 : Composition of polynomials
Dattier   1
N 4 hours ago by Dattier
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
1 reply
Dattier
Yesterday at 1:52 PM
Dattier
4 hours ago
number sequence contains every large number
mathematics2003   3
N 4 hours ago by sttsmet
Source: 2021ChinaTST test3 day1 P2
Given distinct positive integer $ a_1,a_2,…,a_{2020} $. For $ n \ge 2021 $, $a_n$ is the smallest number different from $a_1,a_2,…,a_{n-1}$ which doesn't divide $a_{n-2020}...a_{n-2}a_{n-1}$. Proof that every number large enough appears in the sequence.
3 replies
mathematics2003
Apr 13, 2021
sttsmet
4 hours ago
IMO ShortList 2002, geometry problem 2
orl   28
N 4 hours ago by ezpotd
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
28 replies
orl
Sep 28, 2004
ezpotd
4 hours ago
Arc Midpoints Form Cyclic Quadrilateral
ike.chen   56
N 4 hours ago by ezpotd
Source: ISL 2022/G2
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
56 replies
ike.chen
Jul 9, 2023
ezpotd
4 hours ago
Non-linear Recursive Sequence
amogususususus   4
N 4 hours ago by GreekIdiot
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
4 replies
amogususususus
Jan 24, 2025
GreekIdiot
4 hours ago
Russian Diophantine Equation
LeYohan   2
N 4 hours ago by RagvaloD
Source: Moscow, 1963
Find all integer solutions to

$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$.
2 replies
LeYohan
Yesterday at 2:59 PM
RagvaloD
4 hours ago
Euclid NT
Taco12   12
N Apr 25, 2025 by Ilikeminecraft
Source: 2023 Fall TJ Proof TST, Problem 4
Find all pairs of positive integers $(a,b)$ such that \[ a^2b-1 \mid ab^3-1. \]
Calvin Wang
12 replies
Taco12
Oct 6, 2023
Ilikeminecraft
Apr 25, 2025
Euclid NT
G H J
G H BBookmark kLocked kLocked NReply
Source: 2023 Fall TJ Proof TST, Problem 4
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Taco12
1757 posts
#1 • 2 Y
Y by megarnie, ItsBesi
Find all pairs of positive integers $(a,b)$ such that \[ a^2b-1 \mid ab^3-1. \]
Calvin Wang
This post has been edited 1 time. Last edited by Taco12, Oct 6, 2023, 12:50 AM
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cj13609517288
1923 posts
#2
Y by
The original problem had way more conditions but apparently without the conditions was still solvable and rather interesting lol.

Doing sufficient Euclid yields
\[a^2b-1\mid a^5-1,\]\[a^2b-1\mid a^3-b,\]\[a^2b-1\mid a-b^2.\]
Caseworking on size for the second and third divisibilities yields $\boxed{(1,b)}$, $\boxed{(b^2,b)}$, and $\boxed{(a,a^3)}$, which all work.
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grupyorum
1430 posts
#3
Y by
I claim all answers are $(a,b)=(1,b), (b^2,b), (b,b^3)$, where $b$ is arbitrary.

First, see that for $a=1$, any $b$ works; so let $a>1$. Next, $a^2b-1\le ab^3-1$, so $b^2\ge a$. We have $a^2b-1\mid ab^3-a^2b=ab(b^2-a)$, so $a^2b-1\mid b^2-a$. If $b^2=a$, then any $b$ works, so let $b^2>a$. Now, using $a^2b-1\mid a^3b-a$, we find $a^2b-1 \mid b(a^3-b)$, that is $a^2b-1\mid a^3-b$.
Case 1. $b=a^3$. In this case, clearly any $b$ works.
Case 2. $b>a^3$. Then, $a^2b-1\mid b-a^3$, so $b\ge a^3-1+a^2b\ge 4b$ as $a\ge 2$, a contradiction.
Case 3. $b<a^3$. Then, $a^2b-1\mid a^3-b$, so $a^3\ge a^2b+b-1\ge a^2b$, so $a\ge b$. But then, using $a^2b-1\mid b^2-a$ we have $b^2\ge a+a^2b-1\ge a^2b\ge b^3$, forcingn $b=1$. For $b=1$, $a^2-1\mid a-1$, yielding $a=1$.
This post has been edited 1 time. Last edited by grupyorum, Oct 6, 2023, 1:50 AM
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MathLuis
1555 posts
#4 • 2 Y
Y by vrondoS, MS_asdfgzxcvb
Note that $a^2b-1 \mid ab^3-1 \implies a^2b-1 \mid a^2b^3-a-(a^2b^3-b^2) \implies a^2b-1 \mid b^2-a \implies |b^2-a| \ge a^2b-1$
Also $a^2b-1 \mid ab^3-1 \implies a^2b-1 \mid a^4b^3-a^3 \implies a^2b-1 \mid a^3-b \implies |a^3-b| \ge a^2b-1$.
Now if $a=1$ then all $b$ work so $(1,n)$ is a solution, also note if $b=1$ then $a=1$ (which is $(1,1)$ which we already have), now if $a,b \ge 2$, then we have the following cases.
Case 1: $b \ge a^3$
Then here $b-a^3 \ge a^2b-1$ or $b=a^3$, if the first one holds then $b-8 \ge b-a^3 \ge a^2b-1 \ge 4b-1$ contradiction!, hence another solution pair from this case is $(n,n^3)$
Case 2: $b<a^3$
Case 2.1: $a \ge b^2$
Then $a=b^2$ or $a-b^2 \ge a^2b-1$ hence $a-4 \ge a-b^2 \ge ba^2-1 \ge 2a^2-1$ contradiction! hence $a=b^2$ so $(n^2,n)$ is another solution.
Case 2.2: $b^2>a$
$a^3 \ge a^2b+b-1>a^2b$ hence $a \ge b+1$ hence $b^2-b-1 \ge b^2-a \ge a^2b-1 \ge b^3-2b^2+b-1>b^3-2b^2-b-1$ so $3b^2>b^3$ hence $b=2$ but then $4-a \ge 4a^2-1$ for all $a \ge 3$ which cant happen, so contradiction!.
Hence all the pairs are $(1,n), (n^2, n), (n, n^3)$ thus we are done :D.
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DottedCaculator
7357 posts
#5
Y by
Subtracting gives $a^2b-1\mid ab(b^2-a)$ so $a^2b-1\mid b^2-a$. Therefore, either $a=b^2$, which works, or $a^2b-1\leq b^2-a$. We also have $a^2b-1\mid ab^3-1-(a^4b^2-1)=ab^2(b-a^3)$, so either $b=a^3$, which works, or $a^2b-1\leq a^3-b$. This implies $b<a$, contradicting the first inequality.
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john0512
4190 posts
#7
Y by
The answer is $a=1$ and $a=b^2$ and $b=a^3.$ When $a=1$, the statement is $b-1\mid b^3-1$, which is true by difference of cubes factorization, and when $a=b^2$ the two sides are equal. When $b=a^3$, the statement is $a^5-1\mid a^{10}-1$ which is true by difference of squares.

From now on, assume $a\geq 2$ since we already know that $a=1$ always works.

Rewrite the condition as $$ab^3-1\equiv 0\pmod{a^2b-1}.$$Since $a^2b-1$ is relatively prime to $a$, we will multiply this by $a$ to get $$a^2b^3-a\equiv 0\pmod{a^2b-1}.$$We subtract $a^2b^3-b^2$ from the left side (since that is $b^2$ times the modulus), to get $$b^2\equiv a\pmod{a^2b-1}.$$Now, let $$b^2=a+k(a^2b-1)$$for some integer $k$. If $k=0$, then we have $a=b^2$ which we know works, so from now on assume $k\neq 0.$ Clearly, $k$ cannot be negative either, as $a^2b-1\geq a^2-1>a$ since we are assuming $a\geq 2.$ Thus, $k$ is a positive integer.

Now, rearrange this as a quadratic in $b$ to get $$b^2+(-a^2k)b+k-a=0.$$Since $b$ must be an integer, we have that the discriminant must be a perfect square. Thus, $$a^4k^2+4(a-k)$$must be a square.

Claim 1: $$(a^2k-1)^2<a^4k^2+4(a-k).$$Expanding this out gives $$-2a^2k+1<4a-4k$$$$4k+1<4a+2a^2k.$$This is clearly true, since we are assuming $a\geq 2$ so $$2a^2k+4a\geq 8k+8>4k+1.$$
Claim 2: $$(a^2k+1)^2>a^4k^2+4(a-k).$$Expanding this out gives $$2a^2k+1>4a-4k$$$$2a^2k+4k+1>4a.$$Again, this is clearly true since we are assuming $a\geq 2$ so $$2a^2k+4k+1>2a^2k\geq 2a^2\geq 4a.$$
Thus, the only way for it to be a perfect square is if $$a^4k^2+4(a-k)=(a^2k)^2$$$$a=k.$$
However, if we plug it back into $$b^2+(-a^2k)b+k-a=0,$$this just becomes $$b^2-a^3b=0,$$which has the solution $b=a^3,$ hence done.
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shendrew7
799 posts
#8
Y by
Does $0 \mid 0$? Probably doesn't, so we exclude $(1,1)$ in the solution set.

Note the LHS must be less than or equal to the RHS, so $a \leq b^2$. Euclid also tells us
\[a^2b-1 \mid (ab^3-1)-(a^2b-1) = ab(b^2-a).\]
  • $\min(a,b)=1$: Our solutions are $\boxed{(1,k)}$.
  • Otherwise, $a^2b-1 \mid b^2-a$. If $b \leq a^2$, the RHS must be 0, we get the solution $\boxed{(k^2,k)}$.
  • Otherwise, $a^2b-1 \mid a(a^2b-1) - (b^2-a) = b(b-a^3)$, and as $|b-a^3| < |a^2b-1|$, we require $\boxed{(k,k^2)}$. $\blacksquare$
This post has been edited 1 time. Last edited by shendrew7, Mar 11, 2024, 7:59 PM
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vsamc
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Solution
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RedFireTruck
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Clearly, $a=1$ and $a=b^2$ are both solutions. When $a>b^2$, the LHS is greater than the RHS, so let $1<a<b^2$. Note that $\gcd(a^2b-1, ab^3-1)=\gcd(a^2b-1, b^2-a)$. Clearly, when $a\ge \sqrt{b}$, $b^2-a<a^2b-1$. Therefore, $1<a<\sqrt{b}$ so $1<a^2<b$.

Let $b=a^2+k_2$ for $k_2>0$. Then, $$\gcd(a^2b-1, b^2-a)=\gcd(a^4+k_2a^2-1, a^4+2k_2a^2+k_2^2-a)=\gcd(a^4+k_2a^2-1, k_2a^2+k_2^2-a+1).$$
Clearly, $k_2a^2+k_2^2-a+1>0$, so $k_2a^2+k_2^2-a+1\ge a^4+k_2a^2-1$ so $k_2^2+2\ge a^4+a$. When $k_2=a^2$, $a=2$ so $(2, 8)$ is a solution. Otherwise, $k_2>a^2$. Let $k_2=a^2+k_3$ for $k_3>0$. Then, $$\gcd(a^4+k_2a^2-1, k_2^2+2-a^4-a)=\gcd(2a^4+k_3a^2-1,2a^2k_3+k_3^2-a+2).$$
Clearly, $2a^2k_3+k_3^2-a+2>0$, so $2a^2k_3+k_3^2-a+2\ge 2a^4+k_3a^2-1$, so $a^2k_3+k_3^2+3\ge 2a^4+a$. When $k_3=a^2$, we get $a=3$, so $(3, 27)$ is a solution. Otherwise, $k_3>a^2$. Let $k_3=a^2+k_4$ for $k_4>0$. Then, $$\gcd(2a^4+k_3a^2-1,a^2k_3+k_3^2+3-2a^4-a)=\gcd(3a^4+k_4a^2-1, 3k_4a^2+k_4^2-a+3).$$
Clearly, $3k_4a^2+k_4^2-a+3>0$ so $3k_4a^2+k_4^2-a+3\ge 3a^4+k_4a^2-1$ so we get $2k_4a^2+k_4^2+4\ge 3a^4+a$. When $k_4=a^2$, we get $a=4$ so $(4, 64)$ is a solution. Otherwise, $k_4>a^2$.

It is easy to see that by induction, $k_n=a^2+k_{n+1}$ implies that $((n+1), (n+1)^3)$ is a solution and otherwise $k_{n+1}=a^2+k_{n+2}$. As $b$ must be finite, there are no solutions other than $b=a^3$ in this case.

Therefore, the solutions are $a=1$, $a=b^2$, and $b=a^3$.
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math004
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Let $n=a^2b-1,$ for simplicity, and note that $(a,n)=(b,n)=1.$

\[1\equiv ab^3 \equiv a\times (a^{-2})^3 \equiv a^5 \pmod n  \]So $b$ is just the inverse of $a^2$ modulo $n$ which is $a^3.$
Hence, $b\equiv a^3 \pmod n $ and $b^2\equiv a \pmod n.$

\begin{align*}
     n &\mid b-a^3 \\
     n &\mid b^2-a \\ 
     n &\mid a^5-1   
\end{align*}Now, it is just a size argument and the answer is $(1,n),(n^2,n)$ and $(n,n^3).$
This post has been edited 2 times. Last edited by math004, Jan 26, 2025, 6:56 AM
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pie854
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Notice that $(1,x)$ works. Assume $a, b>1$. Then $a^2b-1 \leq ab^3-1$ i.e. $b^2\geq a$. The pair $(x^2.x)$ works so let's assume $b^2>a$. Note that $$a^2b-1 \mid a(ab^3-1)-b^2(a^2b-1)=b^2-a.$$This implies $b^2+1>a^2b+a$ which implies $b>a^2$. Now $$a^2b-1 \mid b(a^2b-1) - a^2(b^2-a)=a^3-b.$$If $a^3-b>0$ then, $a^3-b\geq a^2b-1$ i.e. $a^3+1 \geq b(a^2+1)>a^2(a^2+1)$, a contradiction. If $a^3-b<0$ then $b-a^3 \geq ab^2-1$ which is clearly absurd. Hence $a^3=b$ and we can check that $(x,x^3)$ works.
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OronSH
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First $a^2b-1\mid -a(ab^3-1)+b^2(a^2b-1)=a-b^2\implies a^2b-1\mid a^2(a-b^2)+b(a^2b-1)=a^3-b$. Now $a\ne 1\implies a^2b-1>b-a^3$ so either $b=a^3$ or $a^3-b\ge a^2b-1\implies a^3-a^2b-b+a\ge 0\implies (a^2+1)(a-b)\ge 0\implies a\ge b$. Now from $a^2b-1\mid a-b^2$ either $a=b^2$ or $a^2b-1\ge b^2-1>b^2-a\ge a^2b-1$ or $a^2b-1>a-1\ge a-b^2\ge a^2b-1$, contradiction. This gives our solutions $(1,t),(t,t^3),(t^2,t)$ which clearly work.
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Ilikeminecraft
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Note that the LHS is relatively prime to $a,$ so we multiply the RHS by $a$. This tells us that $a^2b - 1 \mid a^2b^3 - a \implies a^2b - 1 \mid b^2 - a.$ Multiplying by $a^2,$ it gives $a^2b - 1 \mid b - a^3.$ Now we casework on $b^2$ and $a.$

If $a = b^2,$ this clearly works.

If $a > b^2,$ we have that $a - b^2 < a^2b - 1,$ however, $a - b^2  > 0,$ which gives no solutions.

If $a < b^2,$ we do casework on $a^3 < b$ or $a^3 = b$. Continue by multiplying by $a^2$, and subtracting by the LHS. Clearly, the RHS must be non-negative. Thus, $1 - a^5 = 0 \implies a = 1.$

Thus, the solution set is $(1, k), (k, k^3), (k^2, k)$ where $k > 1.$
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