A diophantine equation

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A diophantine equation

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Source: Turkey Junior National Olympiad 2012 P1

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1345 posts

#1 h Dec 12, 2012, 10:26 AM • 2 Y

Y by Adventure10, Mango247

Let $x, y$ be integers and $p$ be a prime for which

$x^2-3xy+p^2y^2=12p$
Find all triples $(x,y,p)$ .

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488 posts

Babai

#2 h Dec 12, 2012, 1:09 PM • 2 Y

Y by parola, Adventure10

well..write this like $(2x-3y)^2+(4p^2-9)y^2=48p$ ..Separate cases..first with $y=1$ which gives $153-4(p-6)^2$ should be a square..no sol in this case(i checked roughly,.but i thnk its r8)..second case.. $y\geq 2$ ..which gives $48p\geq 4(4p^2-9)$ or $4p^2-12p-9\leq 0$ equivalent to $(2p-3)^2\leq 18<25$ which gives $p<4$ .so $p=2,3$ ..check thiese.. $p=2$ gives $(2x-3y)^2+7y^2=96$ check $y=1,2,3$ is only possible but gives no solution..
$p=3$ ..well the equation is like $(2x-3y)^2+27y^2=144$ ..only possiblity $y=1,2$ .. $y=2$ gives $x=6$ ..So only sol is $(6,2,3)$ ...well again saying checked roughly..there may be some mistakes..but the procedure is same..

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#3 h Dec 13, 2012, 2:53 PM • 5 Y

Y by TBazar, Pluto1708, Adventure10, Mango247, Atilla

Observe that $x^2+p^2y^2 \vdots 3 \Rightarrow x \vdots 3, py \vdots 3$ so the LHS must be divisible by 9, hence the RHS must also divisible by 9, thus we have $p=3$ .
Plug in and solve standard quadratic diophantine equation, we have $(x,y)=(0,2),(0,-2),(6,2),(-6,-2)$

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#4 h Dec 13, 2012, 4:17 PM • 2 Y

Y by Adventure10, Mango247

note that , for integral $x$ , we must have , $y^2(9-4p^2)+48p=z^2$
note that , if $p \neq 2,3$ , then 3 divides $x^2+y^2$ , so , 3 divides x and y
so , 9 divides $[z^2-3p]$ . if $p=1(mod3)$ , then 9 divides $z^2-3$ , which is impossible. similarly , we get a contradiction for $p=-1(mod3)$ . so , we need to check p=2,3 only which is trivial.

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kishi

#5 h Apr 2, 2014, 1:42 PM • 2 Y

Y by Adventure10, Mango247

this question was in my exam in Azerbaijan.....

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#6 h Jun 21, 2015, 1:03 PM • 2 Y

Y by Adventure10, Mango247

crazyfehmy wrote:

Let $x, y$ be integers and $p$ be a prime for which

$x^2-3xy+p^2y^2=12p$
Find all triples $(x,y,p)$ .

The equation can be rewrite as $x(x - 3y) = p(12 - py^2)$ .
if $p = xd$ then $d(pd - 3y) = 12 - py^2$ $\implies$ $p(d^2+y^2) = 3(4 + yd)$ .
If $p = 3$ then $d^2 - yd + y^2 - 4 = 0$ so we get that $16 - 3y^2$ is a perfect square so $y = 2$ or $y = -2$ then $d \in {0,2}$ so $(x,y,p) \in \{(0,2,3),(6,2,3),(0,-2,3),(6,-2,3)\}$ .
if $p$ isn't $3$ then $d,y \equiv 0 ($ $mod$ $3)$ . then $4 \equiv 0 (mod$ $3)$ ,contradiction.
We do similar when $x - 3y = pd$ .

This post has been edited 1 time. Last edited by neverlose, Jun 21, 2015, 1:03 PM

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#7 h Apr 30, 2016, 8:22 PM • 3 Y

Y by dredman, Adventure10, Mango247

Doesn't the triple $(6,0,3)$ work?

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#8 h May 5, 2018, 3:59 PM • 2 Y

Y by dmitar_15, Adventure10

Rewrite as following
$x^2-3xy+p^2y^2-12p=0$

Clearly the discriminant $9y^2-4p^2y^2+48p = d^2$ should be a perfect square.
If $p\neq3$ then $p^2\equiv 1(mod$ $3)$ and

$9y^2-4p^2y^2+48p\equiv -4p^2y^2\equiv 2y^2 \equiv d^2(mod$ $3)$

Since quadratic residues $(mod$ $3)$ are $1$ and $0$ , we can conclude that $3|y$ $\Rightarrow$ $y=3y_1$

$x^2-9xy_1+9p^2y_1^2=12p$ $\Rightarrow$ $3|x$ $\Rightarrow$ $x=3x_1$

$9x_1^2-27x_1y_1+9p^2y_1^2=12p$
$3x_1^2-9x_1y_1+3p^2y_1^2=4p$
$3|LHS$ but not the right side, hence contradiction.

$\Rightarrow$ $p=3$

$x^2-3xy+9y^2=36$ $\Rightarrow 3|x$ $\Rightarrow x=3x_1$
$x_1^2-x_1y+y^2=4$
We solve this quadratic equation by first checking the discriminant and making it be a perfect square and we get the following solutions
$(p,x,y)=(3,6,0),(3,6,2),(3,0,2),(3,0,-2),(3,-6,-2)$

This post has been edited 2 times. Last edited by harapan57, May 5, 2018, 4:06 PM
Reason: typo

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#9 h May 20, 2018, 7:52 PM • 2 Y

Y by Adventure10, Mango247

harapan57 wrote:

Rewrite as following
$x^2-3xy+p^2y^2-12p=0$

Clearly the discriminant $9y^2-4p^2y^2+48p = d^2$ should be a perfect square.
If $p\neq3$ then $p^2\equiv 1(mod$ $3)$ and

$9y^2-4p^2y^2+48p\equiv -4p^2y^2\equiv 2y^2 \equiv d^2(mod$ $3)$

Since quadratic residues $(mod$ $3)$ are $1$ and $0$ , we can conclude that $3|y$ $\Rightarrow$ $y=3y_1$

$x^2-9xy_1+9p^2y_1^2=12p$ $\Rightarrow$ $3|x$ $\Rightarrow$ $x=3x_1$

$9x_1^2-27x_1y_1+9p^2y_1^2=12p$
$3x_1^2-9x_1y_1+3p^2y_1^2=4p$
$3|LHS$ but not the right side, hence contradiction.

$\Rightarrow$ $p=3$

$x^2-3xy+9y^2=36$ $\Rightarrow 3|x$ $\Rightarrow x=3x_1$
$x_1^2-x_1y+y^2=4$
We solve this quadratic equation by first checking the discriminant and making it be a perfect square and we get the following solutions
$(p,x,y)=(3,6,0),(3,6,2),(3,0,2),(3,0,-2),(3,-6,-2)$

Nice solution harapan!

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#10 h Sep 10, 2019, 6:35 AM • 1 Y

Y by Adventure10

harapan57 wrote:

Rewrite as following
$x^2-3xy+p^2y^2-12p=0$

Clearly the discriminant $9y^2-4p^2y^2+48p = d^2$ should be a perfect square.
If $p\neq3$ then $p^2\equiv 1(mod$ $3)$ and

$9y^2-4p^2y^2+48p\equiv -4p^2y^2\equiv 2y^2 \equiv d^2(mod$ $3)$

Since quadratic residues $(mod$ $3)$ are $1$ and $0$ , we can conclude that $3|y$ $\Rightarrow$ $y=3y_1$

$x^2-9xy_1+9p^2y_1^2=12p$ $\Rightarrow$ $3|x$ $\Rightarrow$ $x=3x_1$

$9x_1^2-27x_1y_1+9p^2y_1^2=12p$
$3x_1^2-9x_1y_1+3p^2y_1^2=4p$
$3|LHS$ but not the right side, hence contradiction.

$\Rightarrow$ $p=3$

$x^2-3xy+9y^2=36$ $\Rightarrow 3|x$ $\Rightarrow x=3x_1$
$x_1^2-x_1y+y^2=4$
We solve this quadratic equation by first checking the discriminant and making it be a perfect square and we get the following solutions
$(p,x,y)=(3,6,0),(3,6,2),(3,0,2),(3,0,-2),(3,-6,-2)$

You forgot (3,-6,0)

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#11 h Sep 10, 2019, 8:12 AM • 2 Y

Y by hblanco, Adventure10

$x^2=12p+3xy-p^2y^2\equiv{-p^2y^2}(mod 3)$
Since $p^2y^2\equiv{0 ; 1 }(mod 3)$
So $x^2\equiv{0}(mod 3)$
$\rightarrow{x\equiv{0}(mod 3)}$
$\rightarrow{p^2y^2\equiv{0}(mod 3)}$

Hence we get 2 cases such that :
$(i)$ $y\equiv{0}(mod 3)$
Then $LHS\equiv{0}(mod 9)$
$\Rightarrow{12p\equiv{0}(mod 9)\rightarrow{p=3}}$
$(ii)$ $p\equiv{0}(mod 3)$ implies the same thing that $p=3$ is the only solution .
So the equation becomes :

$x^2-3xy+9y^2=36$
$\Rightarrow{9y^2-3xy+(x^2-36)=0}$
$\Delta=9x^2-36(x^2-36)>0$
Which implies that $36^2>27x^2\rightarrow{48>x^2}$
Don't forget that $x\equiv{0}(mod 3)$
So only candidates are $x\in{(0 ; 3 ; 6 ; -3 ; -6)}$
Testing we see that the solutions are :

$(x;y;p)=(6;0;3)=(-6;0;3)=(0;2;3)=(0;-2;3)=(-6;-2;3)=(6;2;3)$

. $Q.E.D$

This post has been edited 8 times. Last edited by Kamran011, Nov 14, 2019, 6:06 PM

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Atilla

#12 h Dec 12, 2024, 6:46 AM

Y by

Nguyenhuyhoang wrote:

Observe that $x^2+p^2y^2 \vdots 3 \Rightarrow x \vdots 3, py \vdots 3$ so the LHS must be divisible by 9, hence the RHS must also divisible by 9, thus we have $p=3$ .
Plug in and solve standard quadratic diophantine equation, we have $(x,y)=(0,2),(0,-2),(6,2),(-6,-2)$

Good solution bravo easiest in this Propblem l can help to solve quadratic equation
Solution:After puting p=3
We get x²-3xy+(9y²-36)=0
Here ∆=144-27y²
Here 144-27y²must be perfect square for x integer and only y here is y=2,-2
Puting it we found x by ∆ and then we will get the Answer

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sansgankrsngupta

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sansgankrsngupta

#14 h Mar 12, 2025, 9:27 AM

Y by

OG! $mod3$ FINISHES OFF
it directly .

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Primeniyazidayi

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Primeniyazidayi

#15 h Monday at 9:18 PM

Y by

Take mod3 to observe that 3|x^2 + p^2y^2 and because we can't divide x and y forever, we get that p=3.After some quadratic stuff,we get bounds for y and some easy cases give us the result.

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