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Source: 7th European Mathematical Cup , Junior Category, Q2

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sqing

#1 h Dec 25, 2018, 9:56 AM • 4 Y

Y by rightways, Illuzion, Adventure10, Mango247

Find all pairs $(x; y)$ of positive integers such that
$xy | x^2 + 2y -1.$

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#2 h Dec 25, 2018, 1:14 PM • 1 Y

Y by Adventure10

sqing wrote:

Find all pairs $(x; y)$ of positive integers such that
$xy | x^2 + 2y -1.$

Solution: We have $x^2 + 2y - 1\equiv 0 \pmod x\implies 2y - 1\equiv 0 \pmod x$ or $2y - 1 = xz$ for some positive integer $z$ . Now we have $y\mid x+z$ or $\frac{xz+1}{2}\mid x+z\implies xz+1\le 2(x+z) \implies (x-2)(z-2)\le 3.$ For $x = 1$ , $\frac{z+1}{2}\mid z+1$ which is true for any odd natural $z$ . For $x = 3$ , we have $z \le 5$ which on checking gives $z = 1$ or $z = 5$ . For $x = 5$ , $z = 3$ , and for $z=1$ , $x$ is any odd. so we have solutions: $(x,y) = (1,k),(3,2),(3,8),(5,8),(2k-1,k) ~~\text{for any positive integer } k.$

This post has been edited 2 times. Last edited by TheDarkPrince, Dec 25, 2018, 2:58 PM

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Bikey

#3 h Dec 25, 2018, 2:35 PM • 2 Y

Y by danepale, Adventure10

TheDarkPrince wrote:

sqing wrote:

Find all pairs $(x; y)$ of positive integers such that
$xy | x^2 + 2y -1.$

Solution: We have $x^2 + 2y - 1\equiv 0 \pmod x\implies 2y - 1\equiv 0 \pmod x$ or $2y - 1 = xz$ for some positive integer $z$ . Now we have $y\mid x+z$ or $\frac{xz+1}{2}\mid x+z\implies xz+1\le 2(x+z) \implies (x-2)(z-2)\le 3.$ For $x = 1$ , $\frac{z+1}{2}\mid z+1$ which is true for any odd natural $z$ . For $x = 3$ , we have $z \le 5$ which on checking gives $z = 1$ or $z = 5$ . For $x = 5$ , $z = 3$ , so we have solutions: $(x,y) = (1,k),(3,2),(3,8),(5,8) ~~\text{for any positive integer } k.$

O.k but check the pair $(2k-1,k)$

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#4 h Jun 14, 2019, 6:26 PM • 2 Y

Y by itslumi, Adventure10

Let $x^2 + 2y - 1 = kxy$
$D = (ky)^2 - 8y + 4 = s^2 \implies 4( (\frac{ky}{2})^2 -2y + 1) = s^2 \implies (\frac{ky}{2})^2 -2y + 1 = k^2$

If $s$ bigger than y, $(ky)^2 - s^2 \le (y+1)^2 - y^2 = 2y + 1$ so $s\le y-1$ and $(\frac{ky}{2})^2 -2y + 1) \le (y-1)^2 \implies k \le 2$ .
If $k = 2$ we get solutions $(1;k)$ and $(2k-1; k)$ . If $k = 1$ than notice that
$y^2 - 8y + 4 \le (y - 5)^2$ and $(y - 6)^2 \le y^2 - 8y + 4$ for all $y$ more than 8. After checking small cases we get $(3;8)$ and $(5;8)$ .

This post has been edited 4 times. Last edited by Daniil02, Jun 15, 2019, 8:10 AM

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#6 h Nov 21, 2022, 6:46 PM • 2 Y

Y by Tellocan, Aoxz

Since $x|x^2+2y-1 \implies x|2y-1 \implies 2y-1=xk(1) \implies xy|x^2 + xk \implies y|x+k \implies x+k=yt(2)$
By $(1)$ we have $y=\frac{xk+1}{2}$ putting in $(2)$ we get $2(x+k)=t(xk+1)$ .If $t \ge 4$ it is obvious to see that
$LHS<RHS$ .Thus $t=1,2,3$

1. $t=1$
Then $2x+2k=xk+1 \implies (x-2)(k-2)=3 \implies x=3,k=5 ; x=5,k=3 \implies (x,y)=(3,8),(5,8)$

2. $t=2$
Then $x+k=xk+1 \implies (x-1)(k-1)=0$ if $x=1 \implies (x,y)=(1,n)$ , if $k=1 \implies (x,y)=(2n-1,n)$

3. $t=3$
Then $2x+2k=3xk+3 \implies \frac{2}{k} + \frac {2}{x} = 3 + \frac{3}{xk}$ if $x,k \ge 2$ then $LHS<RHS$
Thus at least one of them is equal to $1$ .After checking we will not get new solution...

Hence $(x,y)=(3,8),(5,8),(1,n),(2n-1,n)$ , so we are done :-D

:-D

This post has been edited 1 time. Last edited by Ibrahim_K, Nov 22, 2022, 6:48 AM

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EVKV

#7 h Apr 18, 2025, 1:27 PM

Y by

2y-1=xk
y≤ $x^2$ -1 <x
So k≤2x
putting in the value of y
xk<xk +1≤2x+k≤4x
k≤4
Now k≠4,2 as 2y-1 is odd
k=3 => x≤5 which then can be bashed
k=1 => (2a-1,a) for all naturals a

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