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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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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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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
Classic Diophantine
Adywastaken   4
N 11 minutes ago by mrtheory
Source: NMTC 2024/6
Find all natural number solutions to $3^x-5^y=z^2$.
4 replies
Adywastaken
Today at 3:39 PM
mrtheory
11 minutes ago
Where are the Circles?
luminescent   43
N an hour ago by Amkan2022
Source: EGMO 2022/1
Let $ABC$ be an acute-angled triangle in which $BC<AB$ and $BC<CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and $BQ = BC = CP$. Let $T$ be the circumcenter of triangle $APQ$, $H$ the orthocenter of triangle $ABC$, and $S$ the point of intersection of the lines $BQ$ and $CP$. Prove that $T$, $H$, and $S$ are collinear.
43 replies
luminescent
Apr 9, 2022
Amkan2022
an hour ago
Divisibilty...
Sadigly   0
2 hours ago
Source: Azerbaijan Junior NMO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product divides the sum of their squares.
0 replies
Sadigly
2 hours ago
0 replies
Quadratic system
juckter   35
N 2 hours ago by shendrew7
Source: Mexico National Olympiad 2011 Problem 3
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 - 1 = a_2\]\[ a_2^2 + a_2 - 1 = a_3\]\[\hspace*{3.3em} \vdots \]\[a_{n}^2 + a_n - 1 = a_1\]
35 replies
juckter
Jun 22, 2014
shendrew7
2 hours ago
IMO Shortlist 2012, Geometry 3
lyukhson   75
N 3 hours ago by numbertheory97
Source: IMO Shortlist 2012, Geometry 3
In an acute triangle $ABC$ the points $D,E$ and $F$ are the feet of the altitudes through $A,B$ and $C$ respectively. The incenters of the triangles $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.
75 replies
lyukhson
Jul 29, 2013
numbertheory97
3 hours ago
Diophantine
TheUltimate123   31
N 3 hours ago by SomeonecoolLovesMaths
Source: CJMO 2023/1 (https://aops.com/community/c594864h3031323p27271877)
Find all triples of positive integers \((a,b,p)\) with \(p\) prime and \[a^p+b^p=p!.\]
Proposed by IndoMathXdZ
31 replies
TheUltimate123
Mar 29, 2023
SomeonecoolLovesMaths
3 hours ago
Cyclic ine
m4thbl3nd3r   1
N 4 hours ago by arqady
Let $a,b,c>0$ such that $a^2+b^2+c^2=3$. Prove that $$\sum \frac{a^2}{b}+abc \ge 4$$
1 reply
m4thbl3nd3r
Today at 3:34 PM
arqady
4 hours ago
Non-homogenous Inequality
Adywastaken   7
N 4 hours ago by ehuseyinyigit
Source: NMTC 2024/7
$a, b, c\in \mathbb{R_{+}}$ such that $ab+bc+ca=3abc$. Show that $a^2b+b^2c+c^2a \ge 2(a+b+c)-3$. When will equality hold?
7 replies
Adywastaken
Today at 3:42 PM
ehuseyinyigit
4 hours ago
FE with devisibility
fadhool   2
N 4 hours ago by ATM_
if when i solve an fe that is defined in the set of positive integer i found m|f(m) can i set f(m) =km such that k is not constant and of course it depends on m but after some work i find k=c st c is constant is this correct
2 replies
fadhool
Today at 4:25 PM
ATM_
4 hours ago
Japan MO Finals 2023
parkjungmin   2
N 4 hours ago by parkjungmin
It's hard. Help me
2 replies
parkjungmin
Yesterday at 2:35 PM
parkjungmin
4 hours ago
Iranian geometry configuration
Assassino9931   2
N 4 hours ago by Captainscrubz
Source: Al-Khwarizmi Junior International Olympiad 2025 P7
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$, such that $CD$ is not a diameter of its circumcircle. The lines $AD$ and $BC$ intersect at point $P$, so that $A$ lies between $D$ and $P$, and $B$ lies between $C$ and $P$. Suppose triangle $PCD$ is acute and let $H$ be its orthocenter. The points $E$ and $F$ on the lines $BC$ and $AD$, respectively, are such that $BD \parallel HE$ and $AC\parallel HF$. The line through $E$, perpendicular to $BC$, intersects $AD$ at $L$, and the line through $F$, perpendicular to $AD$, intersects $BC$ at $K$. Prove that the points $K$, $L$, $O$ are collinear.

Amir Parsa Hosseini Nayeri, Iran
2 replies
Assassino9931
Today at 9:39 AM
Captainscrubz
4 hours ago
f(m + n) >= f(m) + f(f(n)) - 1
orl   30
N 5 hours ago by ezpotd
Source: IMO Shortlist 2007, A2, AIMO 2008, TST 2, P1, Ukrainian TST 2008 Problem 8
Consider those functions $ f: \mathbb{N} \mapsto \mathbb{N}$ which satisfy the condition
\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]
for all $ m,n \in \mathbb{N}.$ Find all possible values of $ f(2007).$

Author: Nikolai Nikolov, Bulgaria
30 replies
orl
Jul 13, 2008
ezpotd
5 hours ago
Add d or Divide by a
MarkBcc168   25
N 5 hours ago by Entei
Source: ISL 2022 N3
Let $a > 1$ be a positive integer and $d > 1$ be a positive integer coprime to $a$. Let $x_1=1$, and for $k\geq 1$, define
$$x_{k+1} = \begin{cases}
x_k + d &\text{if } a \text{ does not divide } x_k \\
x_k/a & \text{if } a \text{ divides } x_k
\end{cases}$$Find, in terms of $a$ and $d$, the greatest positive integer $n$ for which there exists an index $k$ such that $x_k$ is divisible by $a^n$.
25 replies
MarkBcc168
Jul 9, 2023
Entei
5 hours ago
Alice and Bob play, 8x8 table, white red black, minimum n for victory
parmenides51   14
N 5 hours ago by Ilikeminecraft
Source: JBMO Shortlist 2018 C3
The cells of a $8 \times 8$ table are initially white. Alice and Bob play a game. First Alice paints $n$ of the fields in red. Then Bob chooses $4$ rows and $4$ columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of $n$ such that Alice can win the game no matter how Bob plays.
14 replies
parmenides51
Jul 22, 2019
Ilikeminecraft
5 hours ago
Pair of multiples
Jalil_Huseynov   63
N Apr 28, 2025 by NerdyNashville
Source: APMO 2022 P1
Find all pairs $(a,b)$ of positive integers such that $a^3$ is multiple of $b^2$ and $b-1$ is multiple of $a-1$.
63 replies
Jalil_Huseynov
May 17, 2022
NerdyNashville
Apr 28, 2025
Pair of multiples
G H J
Source: APMO 2022 P1
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Nguyenhuyhoang
207 posts
#50
Y by
Hi guys, can anyone check if the following solution is correct?

**Step 1: Modulo Analysis and Factorization**

Since $b-1$ is a multiple of $a-1$, we can write $b-1 = k(a-1)$ for some positive integer $k$. This implies
$$b \equiv 1 \pmod{a-1}.$$
**Step 2: Inequality Method: Divisor Bounding**

Since $a^3$ is a multiple of $b^2$, we have $b^2 \mid a^3$. By the divisor bounding inequality, we get
$$b \le a^{\frac{3}{2}}.$$
**Step 3: Combining Modulo Analysis and Inequality**

Combining the results from Step 1 and Step 2, we have:
$$1 \le b \le a^{\frac{3}{2}} \text{ and } b \equiv 1 \pmod{a-1}.$$
This means $b$ must be one of the numbers $1, a, 2a-1, 3a-2, \dots$ up to the largest multiple of $a-1$ less than or equal to $a^{\frac{3}{2}}$.

**Step 4: Casework**

* **Case 1: b = 1**
This case trivially satisfies the conditions for any positive integer $a$.

* **Case 2: b = a**
This case also trivially satisfies the conditions for any positive integer $a$.

* **Case 3: b > a**
In this case, we have $b = l(a-1) + 1$ for some positive integer $l \ge 2$. Since $b^2 \mid a^3$, we can write
$$a^3 = m b^2 = m (l(a-1) + 1)^2$$for some positive integer $m$. Expanding the right side, we get a polynomial in $a$ with leading coefficient $ml^2$. Since the left side has leading coefficient $1$, we must have $ml^2 = 1$. This is impossible since $m$ and $l$ are both positive integers greater than 1.

**Step 5: Product of Two Numbers is a Square and Four Numbers Theorem**

We are left with the case where $a < b < 2a-1$. Since $b^2 \mid a^3$, we can write $a^3 = cb^2$ for some positive integer $c$. By the "Product of Two Numbers is a Square" technique, there exist positive integers $d, p, q$ with $\gcd(p,q) = 1$ such that
$$a = dp^2 \text{ and } c = dq^2.$$
Substituting into $a^3 = cb^2$, we get $d^3p^6 = d^2q^2b^2$, or $dp^6 = q^2b^2$. By the Four Numbers Theorem, there exist positive integers $r, s, t, u$ such that
$$d = rs, \ p^3 = tu, \ q = rt, \ b = su.$$
Since $\gcd(p,q) = 1$, we must have $\gcd(t,s) = 1$. Therefore, $s^2 \mid d$ and $t^2 \mid p^3$. This implies $s \mid r$ and $t \mid u$. Writing $r = vs$ and $u = wt$, we get
$$d = v s^2, \ p^3 = vwt^3, \ q = vst, \ b = su.$$
Substituting into $b-1 = k(a-1)$, we get
$$su - 1 = k(vs^2p^2 - 1) = k(vs^2vwt^3 - 1) = k(v^2s^2wt^3 - 1).$$
This implies $su - 1$ is divisible by $s^2$. However, since $s \ge 1$, we have $su - 1 \equiv -1 \pmod{s^2}$, a contradiction.

**Conclusion**

Therefore, the only solutions are $(a,b) = (a,1)$ and $(a,b) = (a,a)$ for any positive integer $a$.
Z K Y
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Nguyenhuyhoang
207 posts
#51
Y by
Hi guys, can anyone check if the following solution is correct? Many thanks!

**Step 1**

Since $b-1$ is a multiple of $a-1$, we can write $b-1 = k(a-1)$ for some positive integer $k$. This implies
$$b \equiv 1 \pmod{a-1}.$$
**Step 2**

Since $a^3$ is a multiple of $b^2$, we have $b^2 \mid a^3$. By the divisor bounding inequality, we get
$$b \le a^{\frac{3}{2}}.$$
**Step 3**

Combining the results from Step 1 and Step 2, we have:
$$1 \le b \le a^{\frac{3}{2}} \text{ and } b \equiv 1 \pmod{a-1}.$$
This means $b$ must be one of the numbers $1, a, 2a-1, 3a-2, \dots$ up to the largest multiple of $a-1$ less than or equal to $a^{\frac{3}{2}}$.

**Step 4: Casework**

* **Case 1: b = 1**
This case trivially satisfies the conditions for any positive integer $a$.

* **Case 2: b = a**
This case also trivially satisfies the conditions for any positive integer $a$.

* **Case 3: b > a**
In this case, we have $b = l(a-1) + 1$ for some positive integer $l \ge 2$. Since $b^2 \mid a^3$, we can write
$$a^3 = m b^2 = m (l(a-1) + 1)^2$$for some positive integer $m$. Expanding the right side, we get a polynomial in $a$ with leading coefficient $ml^2$. Since the left side has leading coefficient $1$, we must have $ml^2 = 1$. This is impossible since $m$ and $l$ are both positive integers greater than 1.

**Step 5: Product of Two Numbers is a Square and Four Numbers Theorem**

We are left with the case where $a < b < 2a-1$. Since $b^2 \mid a^3$, we can write $a^3 = cb^2$ for some positive integer $c$. By the "Product of Two Numbers is a Square" technique, there exist positive integers $d, p, q$ with $\gcd(p,q) = 1$ such that
$$a = dp^2 \text{ and } c = dq^2.$$
Substituting into $a^3 = cb^2$, we get $d^3p^6 = d^2q^2b^2$, or $dp^6 = q^2b^2$. By the Four Numbers Theorem, there exist positive integers $r, s, t, u$ such that
$$d = rs, \ p^3 = tu, \ q = rt, \ b = su.$$
Since $\gcd(p,q) = 1$, we must have $\gcd(t,s) = 1$. Therefore, $s^2 \mid d$ and $t^2 \mid p^3$. This implies $s \mid r$ and $t \mid u$. Writing $r = vs$ and $u = wt$, we get
$$d = v s^2, \ p^3 = vwt^3, \ q = vst, \ b = su.$$
Substituting into $b-1 = k(a-1)$, we get
$$su - 1 = k(vs^2p^2 - 1) = k(vs^2vwt^3 - 1) = k(v^2s^2wt^3 - 1).$$
This implies $su - 1$ is divisible by $s^2$. However, since $s \ge 1$, we have $su - 1 \equiv -1 \pmod{s^2}$, a contradiction.

**Conclusion**

Therefore, the only solutions are $(a,b) = (a,1)$ and $(a,b) = (a,a)$ for any positive integer $a$.
This post has been edited 1 time. Last edited by Nguyenhuyhoang, May 30, 2024, 7:56 PM
Z K Y
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Ywgh1
139 posts
#53
Y by
apmo 2022 p1

Let $k=\frac{a^3}{b^2}$ then we have that $a-1| k-1$
Which only happens when $a=b=k$ and hence we get that our solutions are $(n,n)$ and $(n,1)$.
This post has been edited 1 time. Last edited by Ywgh1, Aug 27, 2024, 11:54 AM
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AshAuktober
1005 posts
#54
Y by
Case 1: $b<a \implies b = 1, a > 1$, which always works.
Case 2: $b = a$ also always works.
Case 3: $b>a$.
Let $a = dx, b = dy$ with $d = gcd(a, b)$.
The equations reduce to $$y^2 \mid d, dx-1 \mid x-y.$$Now let $d = ky^2$.
Then $$kxy^2 - 1 \mid x-y,$$so $$ky^2x-1 \le y-x \le xy-1$$$$\implies ky^2x \le xy \implies ky \le 1 \implies k = y = 1.$$But then $$b = dy = d \le dx = a,$$contradiction. So the only solutions are $\boxed{(a, b) = (t, 1), (t,t)}$ where $t \in \mathbb{N}$. $\square$
Z K Y
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SomeonesPenguin
128 posts
#55
Y by
Here is a quick solution. :-D

We will show that the only solutions are $(a,b)=(a,1)$ and $(a,b)=(a,a)$. If $b=1$, $a$ can be any positive integer (apart from $1$) so suppose that $b\neq 1$.

Let $a^3=k\cdot b^2$ where $k$ is a positive integer. Now look at this equation mod $a-1$ and keep in mind that $b\equiv 1 \pmod{a-1}$. We get $$k\equiv 1\pmod{a-1}$$
Now we begin our size argument.

Case 1. $k=1$

We have that $a^3=b^2=k^6$ so $a=k^2$ and $b=k^3$. Now we have $$k^2-1\mid k^3-1\iff k+1\mid k^2+k+1\iff k+1\mid 1$$
So we have no solutions.

Case 2. $k \ge a$

If $b=a$ we get a solution so suppose that $b\neq a$. We clearly have $a<b$ and $t=\frac{b-1}{a-1}\ge 2 \Rightarrow b \ge 2a-1$. Now we have $a^3=k\cdot b^2\ge a(2a-1)^2$. After expanding, this implies $$0\ge 3a^2-4a+1$$
And this is a contradiction. $\square$
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alexanderhamilton124
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#56
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If $b = 1$, we are done, so assume not. Let $a^3 = kb^2$, we have $a - 1 \mid a^3 - 1 = kb^2 - 1$, $a - 1 \mid b - 1 \mid kb^2 - kb$, so $a - 1 \mid kb - 1 \implies a - 1 \mid (k - 1)b$. Since $a - 1 \mid b - 1$, $(a - 1, b) = 1 \implies a - 1 \mid k - 1$.

If $k = 1$, we have $a^3 = b^2$, so let $a = x^2$, $b = x^3$. $x^2 - 1 \mid x^3 - 1 \implies x + 1 \mid x^2 + x + 1$, a contradiction.

If $k \geq a$, then $kb^2 > a^3$, unless $b = k = a$, which works, so we're done.
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RedFireTruck
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#57
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Clearly $b=1$ works, so assume $b>1$.

Let $a^3=nb^2$ and $n=c^3d$ where $n,c,d\ge 1$ are integers and $c$ is maximized. Also note that $n\le b$ because $a\le b$. We see that we can let $a=cdk^2$ and $b=dk^3$ where $k\ge 1$ is an integer. Since $n\le b$, this means that $c\le k$. Plugging this in gives $(cdk^2-1)|(dk^3-1)$ so $(cdk^2-1)|(cdk^3-c)$ so $(cdk^2-1)|(k-c)$. Plugging in $k=c$ gives $a=b$, which works. When $c < k$, the LHS is bigger than the RHS.

Therefore, the solutions are $\boxed{b=1}$ and $\boxed{a=b}$.
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Sadigly
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#58 • 1 Y
Y by alexanderhamilton124
Solution by Mhremath & Sadigly

$a-1\mid b-1$ Either $b=1$ or $b\geq a$. For $b=1$ case, $(a;b)=(a;1)$ obviously works

Let $a^3=kb^2$ for a $k\in\mathbb{Z}^+$

Claim. $a-1\mid k-1$
Proof. We have $a-1\mid (b-1)(b+1)k$ or in other words $a-1\mid b^2k-k\Rightarrow a-1\mid a^3-k$. We also have $a-1\mid a^3-1$ for $a\neq1$. Subtracting these two gives us $a-1\mid k-1$

This could mean 3 things.

1. Case $k=1\Rightarrow a^3=b^2=x^6$ for some $x\in\mathbb{Z}^+/\{1\}$
$$a-1\mid b-1\Rightarrow x^2-1\mid x^3-1\Rightarrow x+1\mid x^2+x+1$$But we have $gcd(x+1;x^2+x+1)=1$ So this case can't be true

2. Case $k=a\Rightarrow a=b$ This obviously works.

3. Case $k>a\Rightarrow a^3=kb^2>ab^2\Rightarrow a>b$ But we have $b\geq a$, so this case can't be true,too
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math004
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#59
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Let $k=a^3/b^2.$
\[0\equiv a^3-1 = kb^2-1\equiv k-1 \pmod{a-1}.\]However, $a-1\mid b-1\implies b\leq a \implies k\leq a.$ This forces $k=a,$ in which case $b^2=a^2\implies a=b$ or $k=1.$
When $k=1, a^3=b^2\implies (a,b)=(x^2,x^3).$ $$a-1\mid b-1 \iff x^2-1\mid x^3-1  \iff x+1\mid 1+x+x^2 \iff x+1\mid 1.$$Which is obviously impossible. Convsersely, We can easily verify that $(a,a)$ is a solution.
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pie854
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#60
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Taking $b=1$ works. Suppose $b>1$, then $a-1\mid b-1$ gives $a\leq b$. Note that $b^2(a-1) \mid a^3(b^2-1)$. So, $$b^2a-b^2 \mid -a^3(b^2-1)+(a^2+a+1)(b^2a-b^2)=a^3-b^2.$$Thus, $b^2a-b^2\leq a^3-b^2 \implies b\leq a$ (if $a^3=b^2$ then let $a=x^2$, $b=x^3$ and thus $x^2-1\mid x^3-1$ i.e. $x=1$, a contradiction) and it follows that $a=b$.
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SSS_123
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#61
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Given conditions,
$b^2|a^3$
$a-1|b-1$
where $a,b$ are positive integers.
Let $a,b \neq 1$
From the 2nd condition we have,
$a-1 \leq b-1$
$a \leq b$
From the first condition we have,
$b^2|a^3$
So we can write $a^3=b^2k$ for some positive integer $k$.

Now
$ b-1 \equiv 0 \mod{(a-1)}$
$ b \equiv 1 \mod{(a-1)}$
$ b^2 \equiv 1 \mod{(a-1)}$
$ b^2k \equiv k \mod{(a-1)}$
$ a^3 \equiv k \mod{(a-1)}$
$ 1 \equiv k \mod{(a-1)}$
$ 1  \equiv \frac{a^3}{b^2}  \mod{(a-1)}$
$ \frac{a^3}{b^2} - 1 \equiv 0 \mod{(a-1)}$
So
$a-1|\frac{a^3}{b^2} - 1$
$a-1|\frac{a^3-b^2}{b^2}$
$b^2(a-1)|a^3-b^2$
$ab^2-b^2|a^3-b^2$
$ab^2-b^2 \leq a^3-b^2$
$ab^2 \leq a^3$
$b^2 \leq a^2$
$b \leq a$
But we also got $a \leq b$
Thus $a=b$
Now if $a=1$ then it is obvious that the conditions won't fullfill
if $b=1$,it is easy to see that it works for all positive integers $a$
Thus the required solutions are:
$(a,b)=(x,1),(x,x)$ for all positive integers $x$.
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Iveela
116 posts
#62
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Answer: $(a, b) = (x, 1)$ and $(x, x)$.

Suppose there exist other solutions. Let $b - 1 = k(a - 1)$ which we can rewrite as $b = (a - 1)k + 1$. Then $d = \text{gcd}(a, b) = \text{gcd}(a,  k - 1) \leq k - 1$. Now, notice that $b^2 \mid a^3 \Leftrightarrow \left (\frac{b}{d} \right)^2 \mid \left( \frac{a}{d} \right)^3 \cdot d$ implies $\left( \frac{b}{d} \right)^2 \mid d$. Consequently, we get the following bound for $k$.
\[\frac{b^2}{d^2} \leq d \implies  a^2 \leq b^2 \leq d^3 \leq (k - 1)^3 \implies a^{\frac{2}{3}} + 1 \leq k.\]Plugging this back in, we receive $b \geq (a - 1)(a^{\frac{2}{3}} + 1) + 1 \geq a^{\frac{5}{3}}$ which implies $b^2 > a^3$, a contradiction. $\square$
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ray66
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#63
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$(a,b)=(x,1)$ and $(x,x)$ for a positive integer $x$.

First, verify that $b=1$ works. Next, assume $b \ge 2$. Then $b \ge a$ from $b-1$ is a multiple of $a-1$. Also, $a^3=qb^2$. If $q=a$, then $a=b$. Otherwise, $b^2 \equiv 0 \pmod a$ so $a | b$. Writing $b-1 = k(a-1)$ gives $$ b-1 \equiv -k \pmod a$$or $$k \equiv 1 \pmod a$$. $k=a+1$ gives $b=a^2$, but then $a^4 | a^3$, contradiction. So $k=1$ and $a=b$.
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Ilikeminecraft
627 posts
#64
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By the first condition, we have that $a^3 = kb^2$ for some $k\in\mathbb Z.$ Hence, $a - 1 \mid b - 1 \mid b^2 - 1 \mid kb^2 - k = a^3 - k.$ Hence, we conclude $k\equiv 1\pmod {a - 1}.$ If $k\neq a,$ then $k\geq 2a - 1,$ which is clearly absurd since our second condition implies $b \geq a.$ Thus, $k = a,$ and so $\boxed{a = b > 1}$ is the only solution.
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NerdyNashville
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#65
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Solution
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