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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
2n^2+4n-1 and 3n+4 have common powers
bin_sherlo   4
N a minute ago by CM1910
Source: Türkiye 2025 JBMO TST P5
Find all positive integers $n$ such that a positive integer power of $2n^2+4n-1$ equals to a positive integer power of $3n+4$.
4 replies
bin_sherlo
Yesterday at 7:13 PM
CM1910
a minute ago
An interesting geometry
k.vasilev   19
N 8 minutes ago by Ilikeminecraft
Source: All-Russian Olympiad 2019 grade 10 problem 4
Let $ABC$ be an acute-angled triangle with $AC<BC.$ A circle passes through $A$ and $B$ and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$ respectively. The circumcircles of $A_1B_1C$ and $ABC$ meet each other at points $P$ and $C.$ The segments $AB_1$ and $A_1B$ intersect at $S.$ Let $Q$ and $R$ be the reflections of $S$ in the lines $CA$ and $CB$ respectively. Prove that the points $P,$ $Q,$ $R,$ and $C$ are concyclic.
19 replies
k.vasilev
Apr 23, 2019
Ilikeminecraft
8 minutes ago
Squeezing Between Perfect Squares and Modular Arithmetic(JBMO TST Turkey 2025)
HeshTarg   0
11 minutes ago
Source: Turkey JBMO TST Problem 4
If $p$ and $q$ are primes and $pq(p+1)(q+1)+1$ is a perfect square, prove that $pq+1$ is a perfect square.
0 replies
HeshTarg
11 minutes ago
0 replies
sum (a+b)/(a^2+ab+b^2) <=2 if 1/a+1/b+1/c =3 for a,b,c>0
parmenides51   15
N 11 minutes ago by AylyGayypow009
Source: 2020 Greek JBMO TST p2
Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}=3$. Prove that
$$\frac{a+b}{a^2+ab+b^2}+ \frac{b+c}{b^2+bc+c^2}+ \frac{c+a}{c^2+ca+a^2}\le 2$$When is the equality valid?
15 replies
parmenides51
Nov 14, 2020
AylyGayypow009
11 minutes ago
Bosnia and Herzegovina JBMO TST 2016 Problem 3
gobathegreat   3
N an hour ago by Sh309had
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2016
Let $O$ be a center of circle which passes through vertices of quadrilateral $ABCD$, which has perpendicular diagonals. Prove that sum of distances of point $O$ to sides of quadrilateral $ABCD$ is equal to half of perimeter of $ABCD$.
3 replies
gobathegreat
Sep 16, 2018
Sh309had
an hour ago
Power Of Factorials
Kassuno   180
N an hour ago by maromex
Source: IMO 2019 Problem 4
Find all pairs $(k,n)$ of positive integers such that \[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]Proposed by Gabriel Chicas Reyes, El Salvador
180 replies
1 viewing
Kassuno
Jul 17, 2019
maromex
an hour ago
100 card with 43 having odd integers on them
falantrng   7
N an hour ago by Just1
Source: Azerbaijan JBMO TST 2018, D2 P4
In the beginning, there are $100$ cards on the table, and each card has a positive integer written on it. An odd number is written on exactly $43$ cards. Every minute, the following operation is performed: for all possible sets of $3$ cards on the table, the product of the numbers on these three cards is calculated, all the obtained results are summed, and this sum is written on a new card and placed on the table. A day later, it turns out that there is a card on the table, the number written on this card is divisible by $2^{2018}.$ Prove that one hour after the start of the process, there was a card on the table that the number written on that card is divisible by $2^{2018}.$
7 replies
falantrng
Aug 1, 2023
Just1
an hour ago
GCD and LCM operations
BR1F1SZ   1
N an hour ago by WallyWalrus
Source: 2025 Francophone MO Juniors P4
Charlotte writes the integers $1,2,3,\ldots,2025$ on the board. Charlotte has two operations available: the GCD operation and the LCM operation.
[list]
[*]The GCD operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{gcd}(a, b)$.
[*]The LCM operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{lcm}(a, b)$.
[/list]
An integer $N$ is called a winning number if there exists a sequence of operations such that, at the end, the only integer left on the board is $N$. Find all winning integers among $\{1,2,3,\ldots,2025\}$ and, for each of them, determine the minimum number of GCD operations Charlotte must use.

Note: The number $\operatorname{gcd}(a, b)$ denotes the greatest common divisor of $a$ and $b$, while the number $\operatorname{lcm}(a, b)$ denotes the least common multiple of $a$ and $b$.
1 reply
BR1F1SZ
Saturday at 11:24 PM
WallyWalrus
an hour ago
x^2+4 = y^5
Valentin Vornicu   14
N an hour ago by Rayvhs
Source: Balkan MO 1998, Problem 4
Prove that the following equation has no solution in integer numbers: \[ x^2 + 4 = y^5.  \] Bulgaria
14 replies
Valentin Vornicu
Apr 24, 2006
Rayvhs
an hour ago
2023 Japan Mathematical Olympiad Preliminary
parkjungmin   0
2 hours ago
Please help me if I can solve the Chinese question
0 replies
parkjungmin
2 hours ago
0 replies
Polynomial divisible by x^2+1
Miquel-point   1
N 3 hours ago by luutrongphuc
Source: Romanian IMO TST 1981, P1 Day 1
Consider the polynomial $P(X)=X^{p-1}+X^{p-2}+\ldots+X+1$, where $p>2$ is a prime number. Show that if $n$ is an even number, then the polynomial \[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]is divisible by $X^2+1$.

Mircea Becheanu
1 reply
Miquel-point
Apr 6, 2025
luutrongphuc
3 hours ago
the same prime factors
andria   5
N 3 hours ago by bin_sherlo
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
5 replies
andria
Sep 6, 2015
bin_sherlo
3 hours ago
A sharp estimation of the product
mihaig   0
3 hours ago
Source: VL
Let $n\geq4$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$a_1+\cdots+a_n+2^{n-1}\geq n+2^{n-1}\cdot\prod_{i=1}^{n}{a_i}.$$
0 replies
mihaig
3 hours ago
0 replies
xf(x) + f ^2(y) +2 xf(y) perfect square for all positive integers x,y
parmenides51   9
N 3 hours ago by Gaunter_O_Dim_of_math
Source: Balkan BMO Shortlist 2017 N2
Find all functions $f :Z_{>0} \to Z_{>0}$ such that the number $xf(x) + f ^2(y) + 2xf(y)$ is a perfect square for all positive integers $x,y$.
9 replies
parmenides51
Aug 1, 2019
Gaunter_O_Dim_of_math
3 hours ago
IMO ShortList 1998, number theory problem 1
orl   56
N Yesterday at 5:55 PM by Markas
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 $xy^{2}+y+7$.
56 replies
orl
Oct 22, 2004
Markas
Yesterday at 5:55 PM
IMO ShortList 1998, number theory problem 1
G H J
Source: IMO ShortList 1998, number theory problem 1
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huashiliao2020
1292 posts
#44 • 1 Y
Y by cubres
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
1541 posts
#45 • 1 Y
Y by cubres
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.
Z K Y
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HamstPan38825
8864 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.
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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.
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shendrew7
796 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 $\boxed{(11,1),(49,1)}$. $\blacksquare$
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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. $\blacksquare$
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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
4223 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
1271 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
23 posts
#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
632 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
47 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
150 posts
#58
Y by
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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