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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
IMO Genre Predictions
ohiorizzler1434   46
N 15 minutes ago by Mrcuberoot
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
46 replies
ohiorizzler1434
May 3, 2025
Mrcuberoot
15 minutes ago
4 wise men and 100 hats. 3 must guess their numbers
NO_SQUARES   1
N 21 minutes ago by noemiemath
Source: 239 MO 2025 10-11 p5
There are four wise men in a row, each sees only those following him in the row, i.e. the $1$st sees the other three, the $2$nd sees the $3$rd and $4$th, and the $3$rd sees only the $4$th. The devil has $100$ hats, numbered from $1$ to $100$, he puts one hat on each wise man, and hides the extra $96$ hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
1 reply
+1 w
NO_SQUARES
an hour ago
noemiemath
21 minutes ago
IMO Shortlist 2011, G4
WakeUp   126
N 29 minutes ago by NuMBeRaToRiC
Source: IMO Shortlist 2011, G4
Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$. Prove that the points $D,G$ and $X$ are collinear.

Proposed by Ismail Isaev and Mikhail Isaev, Russia
126 replies
WakeUp
Jul 13, 2012
NuMBeRaToRiC
29 minutes ago
<DPA+ <AQD =< QIP wanted, incircle circumcircle related
parmenides51   42
N 32 minutes ago by AR17296174
Source: IMo 2019 SL G6
Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA + \angle AQD =\angle QIP$.

(Slovakia)
42 replies
parmenides51
Sep 22, 2020
AR17296174
32 minutes ago
Help my diagram has too many points
MarkBcc168   28
N 35 minutes ago by AR17296174
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
28 replies
MarkBcc168
Jul 17, 2024
AR17296174
35 minutes ago
A lot of circles
ryan17   8
N 36 minutes ago by AR17296174
Source: 2019 Polish MO Finals
Denote by $\Omega$ the circumcircle of the acute triangle $ABC$. Point $D$ is the midpoint of the arc $BC$ of $\Omega$ not containing $A$. Circle $\omega$ centered at $D$ is tangent to the segment $BC$ at point $E$. Tangents to the circle $\omega$ passing through point $A$ intersect line $BC$ at points $K$ and $L$ such that points $B, K, L, C$ lie on the line $BC$ in that order. Circle $\gamma_1$ is tangent to the segments $AL$ and $BL$ and to the circle $\Omega$ at point $M$. Circle $\gamma_2$ is tangent to the segments $AK$ and $CK$ and to the circle $\Omega$ at point $N$. Lines $KN$ and $LM$ intersect at point $P$. Prove that $\sphericalangle KAP = \sphericalangle EAL$.
8 replies
ryan17
Jul 9, 2019
AR17296174
36 minutes ago
NT FE from Taiwan TST
Kitayama_Yuji   13
N 41 minutes ago by bin_sherlo
Source: 2024 Taiwan TST Round 2 Mock P3
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f\colon \mathbb{N}\to \mathbb{N}$ such that $mf(m)+(f(f(m))+n)^2$ divides $4m^4+n^2f(f(n))^2$ for all positive integers $m$ and $n$.
13 replies
Kitayama_Yuji
Mar 29, 2024
bin_sherlo
41 minutes ago
Yet another domino problem
juckter   15
N 44 minutes ago by lksb
Source: EGMO 2019 Problem 2
Let $n$ be a positive integer. Dominoes are placed on a $2n \times 2n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
15 replies
juckter
Apr 9, 2019
lksb
44 minutes ago
Difference of counts of any 2 colors in any interesting rectangle is at most 1
NO_SQUARES   0
an hour ago
Source: 239 MO 2025 10-11 p8
Positive integer numbers $n$ and $k > 1$ are given. Losyash likes some of the cells of the $n \times n$ checkerboard. In addition, he is interested in any checkered rectangle with a perimeter of $2n + 2$, the upper-left corner of which coincides with the upper-left corner of the board (there are $n$ such rectangles in total). Given $n$ and $k$, determine whether Losyash can color each cell he likes in one of $k$ colors so that in any rectangle of interest to him the number of cells of any two colors differ by no more than $1$.
0 replies
NO_SQUARES
an hour ago
0 replies
Reflection of H about O and SA + SB + SC + AM < AB + BC + CA if US=UM.
NO_SQUARES   0
an hour ago
Source: 239 MO 2025 10-11 p7
Point $M$ is the midpoint of side $BC$ of an acute—angled triangle $ABC$. The point $U$ is symmetric to the orthocenter $ABC$ relative to its circumcenter. The point $S$ inside triangle $ABC$ is such that $US = UM$. Prove that $SA + SB + SC + AM < AB + BC + CA$.
0 replies
NO_SQUARES
an hour ago
0 replies
If {a^r}={a^s}={a^t}=k, then k=0
NO_SQUARES   0
an hour ago
Source: 239 MO 2025 10-11 p6
The real number $a>1$ is given. Suppose that $r$, $s$ and $t$ are different positive integer numbers such that $\{a^r\}=\{a^s\}=\{a^t\}$. Prove that $\{a^r\}=\{a^s\}=\{a^t\}=0$.
0 replies
NO_SQUARES
an hour ago
0 replies
Flips and flops
NO_SQUARES   0
an hour ago
Source: 239 MO 2025 10-11 p4
The numbers from $1$ to $2025$ are arranged in some order in the cells of the $1 \times 2025$ strip. Let's call a flip an operation that takes two arbitrary cells of a strip and swaps the numbers written in them, but only if the larger of these numbers is located to the left of the smaller one. A flop is a set of several flips that do not contain common cells that are executed simultaneously. (For example, a simultaneous flip between the 2nd and 8th cells and a flip between the 5th and 101st cells.) Prove that there exists a sequence of $66$ flops such that for any initial arrangement, applying this sequence of flops to it will result in the numbers being ordered from left to right in ascending order.
0 replies
NO_SQUARES
an hour ago
0 replies
Geometry with fix circle
falantrng   32
N an hour ago by bjump
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
32 replies
falantrng
Feb 25, 2018
bjump
an hour ago
Problem 6
SlovEcience   4
N an hour ago by ZeltaQN2008
Given two points A and B on the unit circle. The tangents to the circle at A and B intersect at point P. Then:
\[ p = \frac{2ab}{a + b} \], \[ p, a, b \in \mathbb{C} \]
4 replies
SlovEcience
May 3, 2025
ZeltaQN2008
an hour ago
IMO ShortList 1998, number theory problem 1
orl   54
N Apr 25, 2025 by Ilikeminecraft
Source: IMO ShortList 1998, number theory problem 1
Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.
54 replies
orl
Oct 22, 2004
Ilikeminecraft
Apr 25, 2025
IMO ShortList 1998, number theory problem 1
G H J
Source: IMO ShortList 1998, number theory problem 1
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andyxpandy99
365 posts
#42
Y by
I claim the solutions are $(49,1)$, $(11,1)$ and $(7n^2,7n)$ for any positive integer $n$.

First rewrite the divisibility as $$xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7) = y^2-7x$$
Now we have two cases.

Case 1: $y^2-7x \geq 0$.

Note that since $x,y \geq 1$ we clearly have $xy^2+y+7+7x> y^2$ so we must have $y^2- 7x = 0$ or $y^2 = 7x$. If we let $x = 7n^2$ where $n$ is any positive integer then $y = 7n$ and we have our first pair $(7n^2,7n)$.

Case 2: $y^2 - 7x < 0$.

In this case, we are essentially looking at $$xy^2+y+7 \mid 7x-y^2$$Thus we must have $xy^2+y+7 \leq 7x-y^2$ which rearranges to $xy^2+y^2+y+7 \leq 7x$. Note that in order for this to be true, we must have $y \leq 2$ or else $xy^2 > 7x$. Now we simply just check the cases where $y=1$ and $y=2$.

If $y = 1$, then $x+8 \mid 7x-1$ which is equivalent to $x+8 \mid 7(x+8)-(7x-1) = 57$ from which we extract $x = 11$ or $x = 49$. This gives us the pairs $(49,1)$ and $(11,1)$.

If $y = 2$, then $4x+9 \mid 7x-4$ which is equivalent to $4x+9 \mid 7(4x+9)-4(7x-4) = 79$ from which we see there are no positive integer solutions for $x$.

We've exhausted all cases and we're done.
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john0512
4186 posts
#43
Y by
We claim the answer is $(7n^2,7n),(49,1),(11,1).$ These clearly all work. Note that $$x^2y+x+y\equiv 0\pmod{xy^2+y+7}$$$$x^2y^2+xy+y^2\equiv 0\pmod{xy^2+y+7}$$$$y^2-7x\equiv0\pmod{xy^2+y+7}.$$(at the second step we subtracted $x$ times the modulus.) The point of this is that now everything is linear in $x$.

Case 1: $y^2-7x>0$. Then, we have $$y^2-7x<y^2\leq xy^2<xy^2+y+7,$$hence there are no solutions as the expression is positive and smaller than the modulus.
f

Case 2: $y^2-7x=0$. Then, clearly $(x,y)=(7n^2,7n)$, which clearly works.

Case 3: $y^2-7x<0$. Then, we have $$7x-y^2=c(xy^2+y+7).$$We can rearrange this to $$x=\frac{y^2+cy+7c}{7-cy^2}.$$
Since $cy^2\leq 6$, there are not that many cases. If $y=1$, we have $$x=\frac{8c+1}{7-c}.$$For $c=4$ and $c=6$ this gives $x=11$ and $x=49$, and other values of $c$ up to 6 do not give integer $x$. If $y=2$, we have $$x=\frac{9c+4}{7-4c},$$here only $c=1$ is allowed which does not give an integer anyway. Finally, $y\geq3$ is not possible, hence done.
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huashiliao2020
1292 posts
#44
Y by
We have $$0\equiv x^2y+x+y\equiv x^2y^2+xy+y^2-(x(xy^2+y+7))\equiv y^2-7x\pmod{xy^2+y+7}.$$Due to size reasons $y^2-7x\le0$, if it equals 0 then the solution set is $(7n^2,7n)$. If it's <0 we must have $xy^2+y+7\le7x-y^2<7x\implies y\in\{1,2\}$; easy calculation reduces it into $\boxed{(x,y)\in\{(7n^2,7n),(11,1),(49,1)\}.}\blacksquare$

wait im not sure why but why are my sols so short lol on overleaf typing they take up like 10 lines
This post has been edited 1 time. Last edited by huashiliao2020, Sep 4, 2023, 4:14 PM
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YaoAOPS
1540 posts
#45
Y by
Note that this implies that $xy^2 + y + 7$ divides $y(x^2y + x + y) - x(xy^2 + y + 7) = y^2 - 7x$.
This implies that either $y^2 = 7x$, or that $x \ge 2, y^2 \ge 16$ can not hold.
In the first case, we get the solution set $(x, y) = \left(7k^2, 7k\right)$ which can be seen to hold.
We now bash out the remaining cases. It can be seen that if $x = 1$, then there are no solutions.
Then, if $y = 1$, it follows that $x + 8 \mid 1 - 7x$ so $x \in \{11, 49\}$. We can check that $(x, y) = (49, 1)$ and $(x, y) = (11, 1)$ works.
If $y = 2$, then $4x + 9 \mid 4 - 7x$ which implies $4x + 9 \mid 3x - 13$ which has no solutions.
If $y = 3$, then $9x + 10 \mid 9 - 7x$ which has no solutions.
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HamstPan38825
8859 posts
#46
Y by
Rewrite the condition as $$xy^2+y+7 \mid (x^2y^2+xy+7x) - (x^2y^2+xy+y^2) = 7x-y^2.$$Now we have a few cases:

If $y^2 > 7x$, then $y^2 - 7x < y^2 < xy^2+y+7$, which is obviously impossible. If $y^2 < 7x$, then $$xy^2+(y+y^2) + 7 \leq 7x,$$which implies $y^2 \leq 7$ and $y \in \{1, 2\}$. This yields the solutions $(49, 1)$ and $(11, 1)$.

If $y^2 = 7x$, then the entire curve of solutions $(7n^2, 7n)$ can be checked to work.
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kamatadu
480 posts
#47
Y by
Bruh, the first time I solved the problem, I solved it for $x^y + x + y \mid xy^2 + y + 7$ ;-; :stretcher: . Also, these edge cases are so hard for me to find without making sillies.

I claim that the answers are $(x,y)=(11,1)$, $(49,1)$ and $(7n^2,7n)$ for any $n\in\mathbb N$.

Firstly, we have $xy^2 + y + 7 \mid x^2y + x + y \mid x^2y^2+xy+y^2\equiv (x^2y^2+xy+y^2)-x(xy^2 + y + 7) = y^2 - 7x$.

Now if $y^2 > 7x$, then we get that $xy^2 + y + 7 \le y^2 - 7x$ which gives us $y^2 -y(1+x^2) -(7+7x) \ge 0$. But then the discriminant must be $\le 0$, that is $(1+x^2)^2 + 4(7+7x) \le 0$ which clearly has no solution.

Now if $y^2 < 7x$, then we get that $xy^2 + y + 7 \ge y^2 - 7x$. Then we get that $x^2y -7x +(y^2+y+7) \le 0$. This means that the discriminant must be $\ge 0$, that is $7^2 -4y(y^2+y+7)\ge 0 \implies 4y(y^2+y+7)\le 49$. This gives only solution $y=1$ since the left side is strictly increasing and it exceeds $49$ for $y=2$. For $y=1$, from the problem statement we get that $x+1+7 \mid x^2 + x + 1 \equiv (x^2+x+1)-x(x+8) = -7x+1 \equiv (-7x+1)+7(x+8) = 57$. This gives us that $x\in \left\{11,49\right\}$ each of which are solutions.

Now for the other case when $y^2 = 7x$, then clearly $7\mid y$ and so $49n^2 = 7x \implies x = 7n^2$ which is another solution.
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shendrew7
794 posts
#48
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Notice the LHs also divides
\[y(x^2y+x+y)-x(xy^2+y+7) = y^2-7x.\]
If $y^2-7x=0$, we have the solutions $\boxed{(7k^2,7k)}$. Otherwise, we notice
\[|xy^2+y+7| > |y^2-7x|,\]implying there are no solutions, unless $y=1,2$ where we get the pairs $\boxed{(11,1),(49,1)}$. $\blacksquare$
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MagicalToaster53
159 posts
#49
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I claim that all solutions are $\boxed{(a, b) \in \{(11, 1), (49, 1), (7t^2, 7t), \}}$, for $t \in \mathbb{Z}^+$.

Observe that \[ab^2 + b + 7 \mid a^2b + a + b \implies ab^2 + b + 7 \mid b(a^2b + a + b) - a(ab^2 + b + 7) = b^2 - 7a.\]We now split into two separate cases:

Case 1:$(b^2 - 7a = 0).$ Then $b^2 = 7a$, so that $b \equiv 0 \pmod 7$, which in turn gives us $b = 7t$, for some $t \in \mathbb{Z}^+$. Then we find the corresponding $a = 7t^2$. $\square$

Case 2:$(b^2 - 7a < 0).$ First observe that $b^2 - 7a \ngeq 0$, else $b^2 - 7a \geq ab^2 + b + 7 \implies b^2 \geq a(b^2 + 7) + b + 7$, which is a clear contradiction for $a, b > 0$. Hence $b^2 - 7a < 0$, so that we obtain \[ab^2 + b + 7 \leq 7a - b^2 \implies b^2(a + 1) + b + 7 \leq 7a \implies b^2 < 7.\]Hence we split into cases for $b = 1, 2$:

Subcase 2.1: $(b = 1).$ Then \[\frac{7a - 1}{a + 8} \in \mathbb{Z} \implies 7 - \frac{57}{a + 8} \implies a + 8 = 1, 3, 19, 57 \implies \boxed{a = 11, 49}. \bigstar\]
Subcase 2.2: $(b = 2).$ Then \[\frac{7a - 4}{4a + 9} \in \mathbb{Z} \implies 1 + \frac{3a - 13}{4a + 9} \implies a \leq -22, \]which is impossible. Therefore no solution exists in this subcase. $\bigstar$

The only solutions, therefore, are $\boxed{(a, b) \in \{(11, 1), (49, 1), (7t^2, 7t))\}}$, for arbitrary $t \in \mathbb{Z}^+$, as claimed. $\blacksquare$
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Aryan27
40 posts
#50 • 1 Y
Y by GeoKing
The solutions are $(x,y) = (11,1)$, $(49,1)$, $(7k^2,7k)$ for all $k\in\mathbb N$.

Note that, we are given that:
\begin{align*}
xy^2+ y+ 7\mid x^2y + x + y \implies xy^2+ y+7\mid y(x^2y + x + y) - x(xy^2+y+7)= y^2-7x
\end{align*}
Now we divide into cases based on the sign of $y^2-7x$
  • When $y^2-7x> 0$.
    The divisibility condition implies that $y^2-7x\geq xy^2+ y+ 7$
    Clearly, $0<y^2-7x<xy^2+ y+ 7$, contradicting the divisibilty condition.

  • When $y^2-7x=0$.
    in this case we get ,
    $y^2=7x$ , let $y = 7k$ , so$ x = 7k^2$.
    Plugging this back in to the original equation reads:
    \begin{align*}
  343k^4 + 7k + 7 \mid 343k^5 + 7k^2 + 7k 
\end{align*}which is always valid, hence these are always solutions.

  • When $y^2-7x<0$.
    We get:
    \begin{align*}
|y^2-7x|\geq xy^2+ y+ 7 
\implies 7x-y^2\geq xy^2+y+7 \iff x(y^2-7)+y^2+y+7\le 0 \iff y \in \{1,2\}.
\end{align*}
    When $y=1$ we get:
    \begin{align*}
x+8 \mid 7x-1 \iff x+8 \mid 57
\end{align*}This gives $x=11$ and $x=49$.

    When $y=2$
    \begin{align*}
    4x+9 \mid 7x-4\iff 4x+9 \mid 79
\end{align*}which gives no solutions.
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RedFireTruck
4221 posts
#51
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Assume that $\gcd(y, 7)=1$. Then, $$\gcd(xy^2+y+7,x^2y+x+y)=\gcd(xy^2+y+7, y^2-7x)=\gcd(7x^2+y+7, 7x-y^2).$$
We want this to equal $xy^{2}+y+7$, so $7x^2+y+7\ge xy^2+y+7$ so $7x\ge y^2$.

We also want $7x-y^2\ge xy^2+y+7$ or $7(x-1)\ge (x+1)y(y+1)$. This means that $y=1$ or $y=2$. When $y=1$, we get $(x+8)|(x^2+x+1)$ so $(x+8)|57$ so $x=11$ or $x=49$.

When $y=2$, we get $(4x+9)|(2x^2+x+2)$ so $(4x+9)|(4x^2+2x+4)$ so $(4x+9)|(x+22)$ so there are no solutions to $x$.

Now assume that $y=7b$. Then, $7|(xy^{2}+y+7)|(x^{2}y+x+y)$ so $x=7a$. Plugging this in means $(49ab^2+b+1)|(49a^2b+a+b)$. Note that $$\gcd(49ab^2+b+1, 49a^2b+a+b)=\gcd(49ab^2+b+1, b^2-a).$$
Note that $|b^2-a|< 49ab^2+b+1$ so $a=b^2$. Plugging $a=b^2$ back in, we get $(49b^4+b+1)|(49b^5+b^2+b)$, which is always true.

Therefore, the solutions are $(11,1)$, $(49,1)$, and $(7k^2,7k)$ for all positive integer $k$.
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ezpotd
1263 posts
#52
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Observe that we then have $xy^2 + y + 7 \mid x^2y^2 + xy + y^2$, so $xy^2 + y + 7 \mid y^2 - 7x$. We can then divide into cases, if $y^2 > 7x$, then clearly we have no solutions by size. If $y^2 = 7x$, then write $y = 7k, x = 7k^2$, we write $x^2y + x + y = 343k^5 + 7k^2 + 7k, xy^2 + y + 7 = 343k^4 + 7k + 7$, we can see all solutions of this form work. If $y^2  < 7x$, by size we still require $7x - y^2 \ge xy^2 + y + 7$, or equivalently $y^2 < 7$. We then check $y = 2$, we require $4x + 9 \mid 7x - 4$, equivalently $4x + 9 \mid 28x - 16$, equivalently $4x +9 \mid 79$, so there are no solutions for $x$ by divisor analysis. We now check $y = 1$, we require $x + 8\mid 7x - 1$, so we have $x + 8 \mid 57$, so we have $x = 11, 49$. We check $x = 11$ gives $x^2y + x + y = 133$ , $xy^2 + y + 7 = 19$, so this pair works. We then check $x = 49$, we get $x^2y + x + y = 2451, xy^2 + y + 7 = 57$, so this pair works as well. The answers are then $(7k^2, 7k), (49,1), (11,1)$.
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Flint_Steel
38 posts
#53
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\Rightarrow supremacy :wacko:
$ab^2+b+7|a^{2}b+a+b \Rightarrow b(ab+1)+7|ab(ab+1)+b^2 \Rightarrow ab^2+b+7|b^2-7a $
$ ab^2+b+7|ab^2-7a^2  \Rightarrow ab^2+b+7|7a^2+b+7$. Since both sides are positive: $ab^2 \leq 7a^2 \Rightarrow b^2\leq 7a$.
So there is two cases to consider.
First case $b^2=7a$: if we set $b=7k$, then $a=7k^2$, We can easily check that it is a solution with $k$ being a positive integer.
Second case $b^2<7a$: Then from earlier, $ab^2+b+7<7a-b^2 \Rightarrow b^2+b+7<a(7-b^2)$ LHS is positive so RHS should follow. Meaning
$7>b^2$. Then we can manually check and see that $(a,b)=(49,1); (11,1)$ is a solution.
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math004
23 posts
#54
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\[xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x \]which implies that $|y^2-7x| \geq xy^2+y+7$ which gives $x=1$ or $y^2\leq 7.$ or $y^2=7x\implies (x,y)=(7t^2,7t)$ which Convsersely always works. The edge cases give $(1,49)$ and $(1,11)$ as solutions.
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pie854
243 posts
#55
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A bit of long division leads us to $xy^2+y+7 \mid y^2-7x$. Clearly $y^2-7x>0$ isn't possible due to size. If $y^2-7x<0$ then $$7x>7x-y^2>xy^2+y+7>xy^2 \implies y=1,2.$$After checking we find the solution $(11,1),(49,1)$. If $y^2=7x$ then $(x,y)=(7k^2,7k)$ for some $k$, which works.
This post has been edited 1 time. Last edited by pie854, Feb 13, 2025, 9:36 AM
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Ilikeminecraft
614 posts
#56
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We first consider when $(y, 7) = 1.$ Multiply the RHS by $y$ to get $x^2y^2 + xy + y^2,$ and then subtracting $x$ times the LHS, we get $y^2 - 7x.$ Thus, we have that $xy^2 + y + 7 \mid y^2 - 7x.$

If $y^2 > 7x,$ we have that $y^2 - 7x \geq xy^2 + y + 7,$ but this is absurd.

If $7x > y^2,$ we have that $7x - y^2\geq xy^2 + y + 7.$ Thus, $y = 2, 1.$ If $y = 1,$ we have $ x + 8\mid7x - 1.$ Clearly, the solutions are $(11, 1), (49, 1).$ If $y = 2,$ we have that $4x + 9 \mid 7x - 4.$ Multiplying by $4,$ we have that $4x + 9 \mid -79,$ which has no solutions.

Now, we consider $(y, 7) = 7.$ Thus, we have that $y = 7k.$ We have $343xk^2 + 7k + 7 \mid 7x^2k + x + 7k.$ We clearly have that $x = 7m$ for some $m\in\mathbb N.$ Thus, $343mk^2 + k + 1 \mid 343 m^2k + m + k.$ The LHS is very clearly relatively prime to $k,$ so we multiply the RHS by $k$ and then apply Euclids, we have that $343mk^2 + k + 1 \mid k^2-m.$ If it is non-zero, this is clearly absurd. Thus, $m = k^2.$ We get the curve $(7k^2, 7k).$
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