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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
Consecutive squares are floors
ICE_CNME_4   2
N a few seconds ago by BraveHedgehog91

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.
2 replies
ICE_CNME_4
3 hours ago
BraveHedgehog91
a few seconds ago
functional equation with exponentials
produit   6
N 4 minutes ago by produit
Find all solutions of the real valued functional equation:
f(\sqrt{x^2+y^2})=f(x)f(y).
Here we do not assume f is continuous
6 replies
1 viewing
produit
4 hours ago
produit
4 minutes ago
Geomerty
Teacher886699   2
N 7 minutes ago by mickeymouse7133
Source: Geometry plese help Me
To each side aof a convex polygon we assign the maximum area of a triangle contained
in the polygon and having a as one of its sides. Show that the sum of the areas assigned to all
sides of the polygon is not less than twice the area of the polygon.
2 replies
Teacher886699
13 minutes ago
mickeymouse7133
7 minutes ago
Inspired by lbh_qys
sqing   2
N 9 minutes ago by BraveHedgehog91
Source: Own
Let $ a, b $ be real numbers such that $ (a-3)(b-3)(a-b)\neq 0 $ and $ a + b =6 . $ Prove that
$$ \left( \frac{a+k-1}{b - 3} + \frac{b+k-1}{3 - a} + \frac{k+2}{a - b} \right)^2 + 2(a^2 + b^2 )\geq6(k+8)$$Where $ k\in N^+.$
2 replies
sqing
May 20, 2025
BraveHedgehog91
9 minutes ago
2025 KMO Inequality
Jackson0423   0
26 minutes ago
Source: 2025 KMO Round 1 Problem 20

Let \(x_1, x_2, \ldots, x_6\) be real numbers satisfying
\[
x_1 + x_2 + \cdots + x_6 = 6,
\]\[
x_1^2 + x_2^2 + \cdots + x_6^2 = 18.
\]Find the maximum possible value of the product
\[
x_1 x_2 x_3 x_4 x_5 x_6.
\]
0 replies
Jackson0423
26 minutes ago
0 replies
Serbian selection contest for the IMO 2025 - P2
OgnjenTesic   3
N 34 minutes ago by OgnjenTesic
Source: Serbian selection contest for the IMO 2025
Let $ABC$ be an acute triangle. Let $A'$ be the reflection of point $A$ over the line $BC$. Let $O$ and $H$ be the circumcenter and the orthocenter of triangle $ABC$, respectively, and let $E$ be the midpoint of segment $OH$. Let $D$ and $L$ be the points where the reflection of line $AA'$ with respect to line $OA'$ intersects the circumcircle of triangle $ABC$, where point $D$ lies on the arc $BC$ not containing $A$. If \( M \) is a point on the line \( BC \) such that \( OM \perp AD \), prove that \( \angle MAD = \angle EAL \).

Proposed by Strahinja Gvozdić
3 replies
OgnjenTesic
an hour ago
OgnjenTesic
34 minutes ago
JBMO TST- Bosnia and Herzegovina 2022 P1
Motion   5
N an hour ago by justaguy_69
Source: JBMO TST 2022 Bosnia and Herzegovina P1
Let $a,b,c$ be real numbers such that $$a^2-bc=b^2-ca=c^2-ab=2$$. Find the value of $$ab+bc+ca$$and find at least one triplet $(a,b,c)$ that satisfy those conditions.
5 replies
Motion
May 21, 2022
justaguy_69
an hour ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   0
an hour ago
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
0 replies
OgnjenTesic
an hour ago
0 replies
Serbian selection contest for the IMO 2025 - P5
OgnjenTesic   0
an hour ago
Source: Serbian selection contest for the IMO 2025
Determine the smallest positive real number $\alpha$ such that there exists a sequence of positive real numbers $(a_n)$, $n \in \mathbb{N}$, with the property that for every $n \in \mathbb{N}$ it holds that:
\[
        a_1 + \cdots + a_{n+1} < \alpha \cdot a_n.
    \]Proposed by Pavle Martinović
0 replies
OgnjenTesic
an hour ago
0 replies
Serbian selection contest for the IMO 2025 - P4
OgnjenTesic   0
an hour ago
Source: Serbian selection contest for the IMO 2025
For a permutation $\pi$ of the set $A = \{1, 2, \ldots, 2025\}$, define its colorfulness as the greatest natural number $k$ such that:
- For all $1 \le i, j \le 2025$, $i \ne j$, if $|i - j| < k$, then $|\pi(i) - \pi(j)| \ge k$.
What is the maximum possible colorfulness of a permutation of the set $A$? Determine how many such permutations have maximal colorfulness.

Proposed by Pavle Martinović
0 replies
OgnjenTesic
an hour ago
0 replies
Serbian selection contest for the IMO 2025 - P3
OgnjenTesic   0
an hour ago
Source: Serbian selection contest for the IMO 2025
Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that:
- $f$ is strictly increasing,
- there exists $M \in \mathbb{N}$ such that $f(x+1) - f(x) < M$ for all $x \in \mathbb{N}$,
- for every $x \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$ such that
\[
            f(y) = \frac{f(x) + f(x + 2024)}{2}.
        \]Proposed by Pavle Martinović
0 replies
OgnjenTesic
an hour ago
0 replies
Serbian selection contest for the IMO 2025 - P1
OgnjenTesic   0
an hour ago
Source: Serbian selection contest for the IMO 2025
Let \( p \geq 7 \) be a prime number and \( m \in \mathbb{N} \). Prove that
\[\left| p^m - (p - 2)! \right| > p^2.\]Proposed by Miloš Milićev
0 replies
+1 w
OgnjenTesic
an hour ago
0 replies
Upper bound on products in sequence
tapir1729   10
N an hour ago by Mathandski
Source: TSTST 2024, problem 7
An infinite sequence $a_1$, $a_2$, $a_3$, $\ldots$ of real numbers satisfies
\[
a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3}
\]for every positive integer $n$. Prove that there exists a real number $C$ such that $a_{n} a_{n+1} < C$ for every positive integer $n$.

Merlijn Staps
10 replies
1 viewing
tapir1729
Jun 24, 2024
Mathandski
an hour ago
Prove $x+y$ is a composite number.
mt0204   0
an hour ago
Let $x, y \in \mathbb{N}^*$ such that $1000 x^{2023}+2024 y^{2023}$ is divisible by $x+y$ and $x+y>2$. Prove that $x+y$ is a composite number.
0 replies
mt0204
an hour ago
0 replies
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
8867 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.
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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. $\blacksquare$
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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
4226 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
1285 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
51 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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