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Source: KJMO 2022 P8

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47 posts

#1 h Oct 29, 2022, 12:47 PM • 1 Y

Y by rightways

Find all pairs $(x, y)$ of rational numbers such that $xy^2=x^2+2x-3$

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paralellepipedon

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paralellepipedon

#2 h Oct 29, 2022, 3:48 PM • 2 Y

Y by TheHawk, Mango247

Hint

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73 posts

#3 h Jan 2, 2023, 12:07 PM • 2 Y

Y by Mathlion1212, seoneo

A student who participated in scoring(marking) the test papers has given the slip that no one had succeeded getting a single point for this problem.

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414 posts

#4 h Jan 2, 2023, 4:05 PM • 1 Y

Y by ZhuTao

Just a sketch:
I would say $x = \frac{a}{b}$ and $y = \frac{c}{d}$ , with $gcd(a,b) = gcd(c,d) = 1$ and then solve the diophantine equation.
This gives $\frac{ac^2}{bd^2} = \frac{a^2+2ab-3b^2}{b^2}$

or $abc^2 = d^2(a+3b)(a-b)$
Now, as $gcd(a,b) = 1$ , we must have $ab|d^2$ as $ab \nmid (a+3b)(a-b)$
As $gcd(c,d) = gcd(c^2,d^2)$ , we even have $d^2|ab$ , so therefore $d^2=ab$ and $c^2=(a+3b)(a-b)$
$c^2 + (2b)^2 = (a+b)^2$ has to be a pythagorean triplet, and $ab$ has to be a square. We can generate infintely many solutions like this if we fix $ab$ and then see if $(a+3b)(a-b)$ is indeed a square

This post has been edited 1 time. Last edited by straight, Jan 2, 2023, 4:05 PM

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18 posts

#5 h Jan 10, 2023, 1:57 AM • 1 Y

Y by PRMOisTheHardestExam

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Solution.
if $y=0$ , $(x, y)=(1, 0), (-3, 0)$ .
if $y<0$ , it is equal to $y>0$ . WLOG $y>0$ .
$xy^{2}=x^{2}+2x-3$ <=> $x^{2}+(2-y^{2})x-3=0$
This equation only has rational solutions.
$D=(2-y^{2})^{2}+12=k^{2}$ . ( $k\in Q$ )
Let $y=\frac{b}{a}$ . $(a, b>0, a, b\in Z, gcd(a, b)=1)$
$(2-\frac{b^{2}}{a^{2}})^{2}+12=\frac{16a^{4}-4a^{2}b^{2}+b^{4}}{a^{4}}=k^{2}$ .
Therefore, $16a^{4}-4a^{2}b^{2}+b^{4}=(2a)^{4}-(2a)^{2}b^{2}+b^{4}=n^{2}$ . $(n\in Z)$
By Lemma., $b=2a$ , $y=2$ .
Therefore, $(x, y)=(-1, 2), (3, 2)$ .
Beacause of WLOG, $(x, y)=(-1, 2), (-1, -2), (3, 2), (3, -2)$ .
In Conclusion, $(x, y)=(1, 0), (-3, 0), (-1, 2), (-1, -2), (3, 2), (3, -2)$ .

This post has been edited 9 times. Last edited by Mathlion1212, Jan 10, 2023, 2:17 AM
Reason: 2022 KJMO #8

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18 posts

#6 h Jan 10, 2023, 2:18 AM • 1 Y

Y by seojun8978

seojun8978 wrote:

A student who participated in scoring(marking) the test papers has given the slip that no one had succeeded getting a single point for this problem.

I'm a Korean and I took this test.
It's right. I just knew the solution when the test finished...

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5 posts

#7 h Jan 22, 2023, 6:58 AM • 1 Y

Y by Mango247

Mathlion1212 wrote:

Lemma. $x^{4}-x^{2}y^{2}+y^{4}=z^{2}$ 's solution is $(x, y, z)=(n, n, n^{2})$ .

Can you give a proof for this?

This post has been edited 1 time. Last edited by faraday_vij, Jan 22, 2023, 7:01 AM

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526 posts

#8 h Jan 22, 2023, 9:26 AM • 3 Y

Y by faraday_vij, Mango247, Mango247

faraday_vij wrote:

Mathlion1212 wrote:

Lemma. $x^{4}-x^{2}y^{2}+y^{4}=z^{2}$ 's solution is $(x, y, z)=(n, n, n^{2})$ .

Can you give a proof for this?

I know the solution but it is not that trivial
I am on the phone now so I cannot post it
I will post it by tuesday

This post has been edited 1 time. Last edited by Seungjun_Lee, Jan 22, 2023, 9:26 AM

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526 posts

#9 h Jan 23, 2023, 7:35 AM • 3 Y

Y by faraday_vij, Mathlion1212, whwlqkd

Lemma> $x^4 - x^2y^2 + y^4 = z^2$ satisfies if and only if $(x,y,z) = (k,k,k^2)$ or $(0,k,k^2)$ or $(k,0,k^2)$
Proof>
if $xy = 0, (x,y,z) = (k,0,k^2)$ or $(0,k,k^2)$ and if $x=y$ , $(x,y,z) = (k,k,k^2)$
Assuming that $xy >0$ and $x \neq y$
Let $(x,y,z)$ is the solution such that $z$ is the minimum of all the integer solutions
if a prime $p$ such that $p|x, p|y$ exists, $\left(\frac{x}{p},\frac{y}{p},\frac{z}{p^2}\right)$ is also a solution which is contradictory to the minimality of $z$
Therefore, $gcd(x,y)=1$
$x^4 -x^2y^2 + y^4 = z^2 \Leftrightarrow (x^2 - y^2)^2 + (xy)^2 = z^2$
$gcd(x^2 - y^2,xy) = 1$

ⅰ) $x,y$ are all odd numbers
By Pythagorean Triplet, $a,b \in \mathbb{N}$ such that $x^2 - y^2 = 2ab, xy = a^2 - b^2,\,\, gcd(a,b) = 1$ exists
$a^4 - a^2b^2 + b^4 = \left(\frac{x^2+y^2}{2}\right)^2$
Therefore, $\left(a,b,\frac{x^2+y^2}{2}\right)$ also is a solution
$z = (a^2+b^2)^2 > a^4 - a^2b^2 + b^4 = \left(\frac{x^2+y^2}{2}\right)^2$
contradiction

ⅱ)There is an even number in ${x,y}$
$WLOG$ $2|x$
Let $x = 2x_1$
By Pythagorean Triplet, $a,b \in \mathbb{N}$ such that $4x_1^2 - y^2 = a^2 - b^2, 2x_1y = 2ab, \,\, gcd(a,b) = 1$ exists
From $x_1y = ab$ , positive integers $m,n,p,q$ such that $x_1 = mn, y = pq, a = mp, b = nq$ exists
Then $4m^2n^2 - p^2q^2 = m^2p^2 - n^2q^2 \Leftrightarrow n^2(4m^2+q^2) = p^2(m^2 + q^2)$
Since $gcd(x_1,y) = 1, gcd(m,q) = gcd(n,p) = 1$
Since $m^2 + q^2 \not\equiv 0 (mod 3), gcd(4m^2+q^2,m^2+q^2) = 1$
$\therefore m^2 +q^2 = n^2, 4m^2 + q^2 = p^2$

if $2|q$ then $2|p$ , from $m^2 + q^2 = n^2,$
By Pythagorean Triplet, positive integers $u,v$ such that $m=u^2 - v^2, q = 2uv$ and satisfies $u^4 - u^2v^2+v^4 = \left(\frac{p}{2}\right)^2$
$\therefore \left(u,v,\frac{p}{2}\right)$ also is a solution
$z = a^2 + b^2 > a = mp \ge p > \frac{p}{2}$
contradiction

if $q$ is an odd number
By Pythagorean Triplet from $(2m)^2 + q^2 = p^2$ , positive integers $u,v$ such that $2m = 2uv, q = u^2 - v^2$ exists
$u,v$ satisfies $u^4 - u^2v^2 + v^4 = n^2$
$\therefore (u,v,n)$ also is a solution
$z = a^2 + b^2 > b = nq \ge n$
contradiction

Hence, there are no more solutions
$\therefore$ $x^4 - x^2y^2 + y^4 = z^2$ satisfies if and only if $(x,y,z) = (k,k,k^2)$ or $(0,k,k^2)$ or $(k,0,k^2)$

This post has been edited 2 times. Last edited by Seungjun_Lee, Jan 23, 2023, 7:36 AM

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85 posts

#10 h Apr 9, 2025, 12:46 PM

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ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

This post has been edited 2 times. Last edited by mqoi_KOLA, Apr 10, 2025, 2:45 AM

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#11 h Apr 9, 2025, 12:55 PM

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too hard for KJMO lol

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99 posts

#12 h Apr 9, 2025, 1:15 PM

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It was reallllllly hard… I solved P5,6,7 that year, but I cannot even get point in this problem. I heard (almost) no one solved it. It is too hard for KJMO

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