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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
Lemma on tangency involving a parallelogram with orthocenter
Gimbrint   0
a few seconds ago
Source: Own
Let $ABC$ be an acute triangle ($AB<BC$) with circumcircle $\omega$ and orthocenter $H$. Let $M$ be the midpoint of $AC$. Line $BH$ intersects $\omega$ again at $L\neq B$, and line $ML$ intersects $\omega$ again at $P\neq L$. Points $D$ and $E$ lie on $AB$ and $BC$ respectively, such that $BEHD$ is a parallelogram.

Prove that $BP$ is tangent to the circumcircle of triangle $BDE$.
0 replies
Gimbrint
a few seconds ago
0 replies
Consecutive squares are floors
ICE_CNME_4   10
N 5 minutes ago by JARP091

Determine how many positive integers \( n \) have the property that both
\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]are consecutive perfect squares.
10 replies
ICE_CNME_4
Yesterday at 1:50 PM
JARP091
5 minutes ago
Sharygin 2025 CR P10
Gengar_in_Galar   2
N 30 minutes ago by Kappa_Beta_725
Source: Sharygin 2025
An acute-angled triangle with one side equal to the altitude from the opposite vertex is cut from paper. Construct a point inside this triangle such that the square of the distance from it to one of the vertices equals the sum of the squares of distances to to the remaining two vertices. No instruments are available, it is allowed only to fold the paper and to mark the common points of folding lines.
Proposed by: M.Evdokimov
2 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
30 minutes ago
inequality thing
BinariouslyRandom   1
N 38 minutes ago by lbh_qys
Source: Philippine MO 2025 P5
Find the largest real constant $k$ for which the inequality \[ (a^2+3)(b^2+3)(c^2+3)(d^2+3) + k(a-1)(b-1)(c-1)(d-1) \ge 0 \]holds for all real numbers $a$, $b$, $c$, and $d$.

answer
1 reply
BinariouslyRandom
an hour ago
lbh_qys
38 minutes ago
Sharygin 2025 CR P7
Gengar_in_Galar   4
N 38 minutes ago by Kappa_Beta_725
Source: Sharygin 2025
Let $I$, $I_{a}$ be the incenter and the $A$-excenter of a triangle $ABC$; $E$, $F$ be the touching points of the incircle with $AC$, $AB$ respectively; $G$ be the common point of $BE$ and $CF$. The perpendicular to $BC$ from $G$ meets $AI$ at point $J$. Prove that $E$, $F$, $J$, $I_{a}$ are concyclic.
Proposed by:Y.Shcherbatov
4 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
38 minutes ago
Sharygin 2025 CR P12
Gengar_in_Galar   8
N an hour ago by Kappa_Beta_725
Source: Sharygin 2025
Circles $\omega_{1}$ and $\omega_{2}$ are given. Let $M$ be the midpoint of the segment joining their centers, $X$, $Y$ be arbitrary points on $\omega_{1}$, $\omega_{2}$ respectively such that $MX=MY$. Find the locus of the midpoints of segments $XY$.
Proposed by: L Shatunov
8 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
an hour ago
Sharygin 2025 CR P17
Gengar_in_Galar   6
N an hour ago by Kappa_Beta_725
Source: Sharygin 2025
Let $O$, $I$ be the circumcenter and the incenter of an acute-angled scalene triangle $ABC$; $D$, $E$, $F$ be the touching points of its excircle with the side $BC$ and the extensions of $AC$, $AB$ respectively. Prove that if the orthocenter of the triangle $DEF$ lies on the circumcircle of $ABC$, then it is symmetric to the midpoint of the arc $BC$ with respect to $OI$.
Proposed by: P.Puchkov,E.Utkin
6 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
an hour ago
Sharygin 2025 CR P21
Gengar_in_Galar   4
N an hour ago by Kappa_Beta_725
Source: Sharygin 2025
Let $P$ be a point inside a quadrilateral $ABCD$ such that $\angle APB+\angle CPD=180^{\circ}$. Points $P_{a}$, $P_{b}$, $P_{c},$ $P_{d}$ are isogonally conjugated to $P$ with respect to the triangles $BCD$, $CDA$, $DAB$, $ABC$ respectively. Prove that the diagonals of the quadrilaterals $ABCD$ and $P_{a}P_{b}P_{c}P_{d}$ concur.
Proposed by: G.Galyapin
4 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
an hour ago
Sharygin 2025 CR P18
Gengar_in_Galar   6
N an hour ago by Kappa_Beta_725
Source: Sharygin 2025
Let $ABCD$ be a quadrilateral such that the excircles $\omega_{1}$ and $\omega_{2}$ of triangles $ABC$ and $BCD$ touching their sides $AB$ and $BD$ respectively touch the extension of $BC$ at the same point $P$. The segment $AD$ meets $\omega_{2}$ at point $Q$, and the line $AD$ meets $\omega_{1}$ at $R$ and $S$. Prove that one of angles $RPQ$ and $SPQ$ is right
Proposed by: I.Kukharchuk
6 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
an hour ago
Problem 3
EthanWYX2009   5
N 2 hours ago by parkjungmin
Source: 2023 China Second Round P3
Find the smallest positive integer ${k}$ with the following properties $:{}{}{}{}{}$If each positive integer is arbitrarily colored red or blue${}{}{},$
there may be ${}{}{}{}9$ distinct red positive integers $x_1,x_2,\cdots ,x_9,$ satisfying
$$x_1+x_2+\cdots +x_8<x_9,$$or there are $10{}{}{}{}{}{}$ distinct blue positive integers $y_1,y_2,\cdots ,y_{10}$ satisfiying
$${y_1+y_2+\cdots +y_9<y_{10}}.$$
5 replies
EthanWYX2009
Sep 10, 2023
parkjungmin
2 hours ago
Inspired by old results
sqing   1
N 2 hours ago by sqing
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
Made from a well-known result
m4thbl3nd3r   0
2 hours ago
1. Let $a,b,c>0$ such that $$\sqrt{(a+b)(a+c)}+\sqrt{(b+a)(b+c)}+\sqrt{(c+a)(c+b)}=3+a+b+c.$$Prove that $$\sqrt{\frac{a+b}{2}}+\sqrt{\frac{b+c}{2}}+\sqrt{\frac{c+a}{2}}\ge ab+bc+ca.$$2. Let $x,y,z$ be sidelengths of a triangle such that $$x^2+y^2+z^2+6=2(xy+yz+zx).$$Prove that $$2\sqrt{2x}+2\sqrt{2y}+2\sqrt{2z}+(x-y)^2+(y-z)^2+(z-x)^2\ge x^2+y^2+z^2.$$
0 replies
m4thbl3nd3r
2 hours ago
0 replies
Interesting inequalities
sqing   5
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(kbc+bd+cd)  \leq \frac{64k}{27}$$$$a (b+c) (kb c+  b d+  c d) \leq \frac{27k}{4}$$Where $ k\geq 2. $
5 replies
sqing
Yesterday at 12:44 PM
sqing
2 hours ago
Long and wacky inequality
Royal_mhyasd   5
N 3 hours ago by Royal_mhyasd
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
5 replies
Royal_mhyasd
May 12, 2025
Royal_mhyasd
3 hours ago
IMO ShortList 1998, number theory problem 1
orl   58
N May 18, 2025 by MihaiT
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 $xy^{2}+y+7$.
58 replies
orl
Oct 22, 2004
MihaiT
May 18, 2025
IMO ShortList 1998, number theory problem 1
G H J
Source: IMO ShortList 1998, number theory problem 1
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HamstPan38825
8868 posts
#46 • 1 Y
Y by cubres
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.
Z K Y
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kamatadu
480 posts
#47 • 1 Y
Y by cubres
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.
Z K Y
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shendrew7
799 posts
#48 • 1 Y
Y by cubres
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 $\boxed{(11,1),(49,1)}$. $\blacksquare$
Z K Y
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MagicalToaster53
159 posts
#49 • 1 Y
Y by cubres
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. $\blacksquare$
Z K Y
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Aryan27
40 posts
#50 • 2 Y
Y by GeoKing, cubres
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
4243 posts
#51 • 1 Y
Y by cubres
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
1286 posts
#52 • 1 Y
Y by cubres
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 • 1 Y
Y by cubres
\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
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#54 • 1 Y
Y by cubres
\[xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x \]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 • 1 Y
Y by cubres
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
658 posts
#56 • 1 Y
Y by cubres
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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reni_wee
52 posts
#57 • 1 Y
Y by cubres
\begin{align*}
xy^2 + y + 7 & \mid x^2y + x + y \\ 
\implies xy^2 + y + 7 & \mid x(x^2y + x + y) - y(xy^2 + y + 7) \\ 
\implies xy^2 + y + 7 & \mid y^2 -7x \\ 
\end{align*}We now proceed to solve this problem using 3 cases.

Case i. $y^2 - 7x > 0$
$$\implies y^2 -7x \geq xy^2 + y + 7$$As $x$ and $y$ are positive integers,
$y^2 -7x < y^2 < xy^2 + y + 7$
Hence a contradiction.

Case ii. $y^2 - 7x = 0$
$$\implies y^2 = 7x$$$\therefore (x,y) = (7k^2,7k)$ ; $k \in \mathbb{Z^+}$ work.

Case iii. $y^2 - 7x < 0$
$$\implies 7x -y^2 \geq xy^2 + y + 7$$For $y^2 > 7$ we have,
$$7x - y^2 < 7x  < xy^2 + y + 7$$which is a contradiciton.
Hence $y^2 \leq 7 \implies y \leq 2$. Therefore we only need to consider the cases where $y=1$ and $y = 2$

When $y = 1$,
\begin{align*}
x+8 & \mid 7x -1 \\
\implies x+8 & \mid 7(x +8) - (7x-1) \\
\implies x+8 & \mid 57
\end{align*}Hence, $(x,y) = (11,1), (49,1)$ works.

When $y=2$,
\begin{align*}
4x+9 & \mid 7x -4 \\
\implies 4x+9 & \mid 7(4x +9) - 4(7x-4) \\
\implies 4x+9 & \mid 79
\end{align*}$\implies x = 17.5 \not \in \mathbb{Z^+}$.

Therefore the only solutions are $(11,1), (49,1)$ and $(7k^2, 7k)$ ; $k \in \mathbb{Z^+}$
This post has been edited 2 times. Last edited by reni_wee, May 6, 2025, 5:19 PM
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Markas
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#58
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We have that $xy^2 + y + 7 \mid x^2y + x + y$. Now we get that $xy^2 + y + 7 \mid y(x^2y + x + y) - x(xy^2 + y + 7)$ $\Rightarrow$ $xy^2 + y + 7 \mid y^2 - 7x$.

Case 1: $y^2 - 7x > 0$ $\Rightarrow$ we want $xy^2 + y + 7 < y^2 - 7x$ $\Rightarrow$ $xy^2 + y + 7 + 7x < y^2$ - which is impossible $\Rightarrow$ we don't have solutions in this case.

Case 2: $y^2 - 7x < 0$ $\Rightarrow$ $xy^2 + y + 7 \leq 7x - y^2$ $\Rightarrow$ $(x + 1)y^2 + y + 7 \leq 7x$. Now if $y \geq 3$ we have that $9(x + 1) \leq 7x$, which is impossible. If y = 1, then $x + 8 \mid 1 - 7x$ or $x + 8 \mid 57$ $\Rightarrow$ we get the solutions (x,y) = (49,1); (11,1) If y = 2, then $4x + 9 \mid 4 - 7x$ and $4x + 9 \mid 4(4 - 7x) + 7(4x + 9)$ $\Rightarrow$ $4x + 9 \mid 79$ and we don't have any solutions here.

Case 3: $y^2 = 7x$ $\Rightarrow$ $7 \mid y$, let $y = 7k$ $\Rightarrow$ $x = 7k^2$ $\Rightarrow$ all $(x,y) = (7k^2,7k)$ work. We have to check this tho and $49k^4 + 7k + 7 \mid 49k^2 - 49k^2$ which is true $\Rightarrow$ all solutions are $(x,y) = (49,1); (11,1); (7k^2,7k)$.
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zhoujef000
322 posts
#59 • 1 Y
Y by MihaiT
replace $x$ and $y$ with $a$ and $b$ since i was given this with $a$ and $b$ and am too lazy to change everything

We claim the answer is $\boxed{(11,1)},$ $\boxed{(49,1)},$ and $\boxed{(7c^2, 7c)}$ for all positive integers $c.$ We proceed as follows.

Firstly, we show these work.

Observe that $11\cdot1^2+1+7=19\mid 19\cdot 7=133=11^2\cdot 1+11+1,$ and $49\cdot 1^2+1+7=57\mid 57\cdot 43=2451=49^2\cdot1+49+1,$ so $(11,1)$ and $(49,1)$ work. Also, for all positive integers $c,$ $7c^2\cdot(7c)^2+7c+7=343c^4+7c+7\mid (343c^2+7c+7)c=343c^5+7c^2+7c=(7c^2)^2(7c)+7c^2+7c,$ so $(7c^2,7c)$ works for all positive integers $c.$ We now show there are no more solutions.

Since $ab^2+b+7\mid a^2b+a+b,$ we have that $b(a^2b+a+b)-a(ab^2+b+7)=a^2b^2+ab+b^2-(a^2b^2+ab+7a)=b^2-7a\equiv 0\pmod{ab^2+b+7}.$ If $b^2-7a>0,$ then since $a\geq 1,$ $ab^2+b+7>b^2>b^2-7a>0,$ so $b^2-7a\not\equiv 0\pmod{ab^2+b+7},$ a contradiction. Therefore, $b^2-7a\leq 0$ and $7a-b^2\geq 0.$

If $7a-b^2=0,$ then $a=\dfrac{b^2}{7},$ so $7\mid b$ and $b=7c$ for some integers $c.$ Then, $a=\dfrac{b^2}{7}=\dfrac{49c^2}{7}=7c^2,$ so $(a,b)=(7c^2,7c).$ Thus, all of the solutions in this case are of this form.

If $7a-b^2>0,$ then since $ab^2+b+7\mid 7a-b^2,$ we must have $ab^2+b+7\leq 7a-b^2.$ However, if $b\geq 3,$ then $ab^2+b+7\geq 9a+10>7a-b^2,$ so $b=1$ or $b=2.$ If $b=1,$ then $a+8=ab^2+b+7\mid 7a-b^2=7a-1,$ so $a+8\mid 7(a+8)-(7a-1)=57.$ Since $a$ is a positive integer, $a+8=19$ or $a+8=57,$ so $a=11$ or $a=49,$ yielding $(a,b)=(11,1)$ or $(a,b)=(49,1).$

If $b=2,$ then $4a+9=ab^2+b+7\mid 7a-b^2=7a-4.$ However, since $2(4a+9)=8a+18>7a-4>0,$ we must have $4a+9=7a-4,$ so $a=\dfrac{13}{3},$ which is not an integer. Thus, there are no solutions in this case.

As such, the only solutions are $(11,1),$ $(49,1),$ and $(7c^2, 7c)$ for all positive integers $c,$ as desired. $\Box$
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MihaiT
763 posts
#60
Y by
zhoujef000 wrote:
replace $x$ and $y$ with $a$ and $b$ since i was given this with $a$ and $b$ and am too lazy to change everything

We claim the answer is $\boxed{(11,1)},$ $\boxed{(49,1)},$ and $\boxed{(7c^2, 7c)}$ for all positive integers $c.$ We proceed as follows.

Firstly, we show these work.

Observe that $11\cdot1^2+1+7=19\mid 19\cdot 7=133=11^2\cdot 1+11+1,$ and $49\cdot 1^2+1+7=57\mid 57\cdot 43=2451=49^2\cdot1+49+1,$ so $(11,1)$ and $(49,1)$ work. Also, for all positive integers $c,$ $7c^2\cdot(7c)^2+7c+7=343c^4+7c+7\mid (343c^2+7c+7)c=343c^5+7c^2+7c=(7c^2)^2(7c)+7c^2+7c,$ so $(7c^2,7c)$ works for all positive integers $c.$ We now show there are no more solutions.

Since $ab^2+b+7\mid a^2b+a+b,$ we have that $b(a^2b+a+b)-a(ab^2+b+7)=a^2b^2+ab+b^2-(a^2b^2+ab+7a)=b^2-7a\equiv 0\pmod{ab^2+b+7}.$ If $b^2-7a>0,$ then since $a\geq 1,$ $ab^2+b+7>b^2>b^2-7a>0,$ so $b^2-7a\not\equiv 0\pmod{ab^2+b+7},$ a contradiction. Therefore, $b^2-7a\leq 0$ and $7a-b^2\geq 0.$

If $7a-b^2=0,$ then $a=\dfrac{b^2}{7},$ so $7\mid b$ and $b=7c$ for some integers $c.$ Then, $a=\dfrac{b^2}{7}=\dfrac{49c^2}{7}=7c^2,$ so $(a,b)=(7c^2,7c).$ Thus, all of the solutions in this case are of this form.

If $7a-b^2>0,$ then since $ab^2+b+7\mid 7a-b^2,$ we must have $ab^2+b+7\leq 7a-b^2.$ However, if $b\geq 3,$ then $ab^2+b+7\geq 9a+10>7a-b^2,$ so $b=1$ or $b=2.$ If $b=1,$ then $a+8=ab^2+b+7\mid 7a-b^2=7a-1,$ so $a+8\mid 7(a+8)-(7a-1)=57.$ Since $a$ is a positive integer, $a+8=19$ or $a+8=57,$ so $a=11$ or $a=49,$ yielding $(a,b)=(11,1)$ or $(a,b)=(49,1).$

If $b=2,$ then $4a+9=ab^2+b+7\mid 7a-b^2=7a-4.$ However, since $2(4a+9)=8a+18>7a-4>0,$ we must have $4a+9=7a-4,$ so $a=\dfrac{13}{3},$ which is not an integer. Thus, there are no solutions in this case.

As such, the only solutions are $(11,1),$ $(49,1),$ and $(7c^2, 7c)$ for all positive integers $c,$ as desired. $\Box$

beautiful! :first:
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