Question

Hi everybody! Find all positive integral solutions (x,y) of the polynomial equation:
4x^3+4x^2y-15xy^2-18y^3-12x^2+6xy+36y^2+5x-10y=0

Answer 1

any thoughts/ideas?

Answer 2

couldn't u find a longer eq? :P

Answer 3

its a monstrous beast...

Answer 4

:P

Answer 5

I think it is smth like (ax-by)^3 bu i can't tell wht r a and b

Answer 6

but*

Answer 7

good try angela..but u never know that will factorize completely to 3 linear factors!...
Infact this BEAST can be factored to any one of the following:
1.3 linears.
2.1 linear and 1 conic

Answer 8

(x-2*y)*(2*x-1+3*y)*(2*x-5+3*y)

Answer 9

(x-2y)(2x+3y-5)(2x+3y-1)=0

Answer 10

i'm not planning to deal with this..u do it...:P

Answer 11

i got it first, please medal

Answer 12

u rock cantorset! but we cant use comps!

Answer 13

how do you know i use computer

Answer 14

:)

Answer 15

@contorset: lol, its obvious. :P

Answer 16

hey how did u do tht?

Answer 17

well witout at computer its LUCK..

Answer 18

I didn't do it on my own, wolframalpha

Answer 19

maple

Answer 20

and how do u do it in comp?

Answer 21

haha... if u had a good luck u wud put x=2y and inspect!!!

Answer 22

you can do it in wolframalpha,

Answer 23

then u get (x-2y) and get others by dividing

Answer 24

i should not give out my secrets though

Answer 25

k. but tat doesnt solve the problem..

Answer 26

x=2y ? you mean substitute

Answer 27

@saubhik: You need a VERY good luck for that!

Answer 28

yes...

Answer 29

but see tat doesnt..solve the problem

Answer 30

we have to find positive integer solutions..

Answer 31

woww

Answer 32

but you can pick anything, x = 4y for instance

Answer 33

and i thought u guys were smart :P:P:P

Answer 34

i dont see any logic in this problem

Answer 35

ya cantorset that's truly possible but as i said we need luck and brain-breaking insights...

Answer 36

see..now how to find the solution?

Answer 37

for the first one(x-2y),
x = 2I
y=I

Answer 38

u can have infinite pairs (2n,n) for a positive integer n since x-2y is a factor...what for the other two?

Answer 39

yes

Answer 40

what for 2x+3y-5=0

Answer 41

its obvious (1,1) but are there others..THIS IS DIOPHANTINE!!

Answer 42

OK lets see, x + 3y/2 =5/2
x=(5-3y)/2

5-3y should be even,

Answer 43

so y is odd.

Answer 44

what about 2x+3y=1? yay...theres no positive integers here in this equation..so this is infertile:P

Answer 45

Why do you think so?

Answer 46

how can u get postive integers (x,y) from 2x+3y<5

Answer 47

there are none...:P

Answer 48

infertile..mother has no babies:P

Answer 49

that can be solved with euclids algorithm

Answer 50

Oh sorry, i dint read that 'positive' part! :P

Answer 51

this mother 2x+3y=5 has only one baby i.e. (1,1)

Answer 52

So our answer: (1,1) and (2n,n)

Answer 53

OMG!!! This is giving me headache only by reading ur explanations imagine if i had to solve this....:@@@@@@@@

Answer 54

So its done, (n,2n) U (1,1)

Answer 55

Openstudy rocks!! thnx evrybody!

Answer 56

Oops (2n,n).

yes right

Answer 57

Going to deal with literature bye and have a good time :)

Answer 58

, amog, you wrote x= 1, y = 21?

Answer 59

@angela: buh-bye.

@cantorset: It was an 'I' (eye)

Answer 60

for the first one(x-2y), x = 2I y=I

Answer 61

Yes i corrected it, didn't I?

Answer 62

hi cantorset...do u mind givin me an idea how to have a factor to the eqn..

Answer 63

suppose i have the baddest luck and go on inspecting from x=3y;x=4y....

Answer 64

wait, ... start over . why would i think to use x = 3y or anything like that

Answer 65

also its 3 linear, not a conic

Answer 66

or worser x=3y+7 etc..

Answer 67

yeah

Answer 68

you mean because the solution is ( ax + by ) ( dx + cy + ...

Answer 69

See cantormath..the given equation
Infact this BEAST can be factored to any one of the following:
1.3 linears.
2.1 linear and 1 conic

Answer 70

now clearly theres no constant term in this equation...so a linear factor (which must be present) must pass through the origin...

Answer 71

and so we start checking by x=y;x=2y;x=3y...

Answer 72

if the question does not mean to torment us we can get the factor in a few trials...

Answer 73

k..thnx evrybody for helping..:)

Answer 74

wait, youre saying in general, a cubic in x and y can be factored into either 3 linears, or 1 linear and 2 conics? , or 2 linears, and a conic?

Answer 75

what do you mean by constant term in the equation

Answer 76

term with no variables attched..

Answer 77

oh right

Answer 78

the final solution is 3 linears, right?

Answer 79

yups...family of 3 straight line

Answer 80

cool, so ok , theres no constant term, and clearly, (0,0) works

Answer 81

yups..try looking for x=y,2y,3y,...here u need luck

Answer 82

so that tells us that x = ay or y = ax must be one of the solutions?

Answer 83

yes..true

Answer 84

but only for a=2

Answer 85

i.e.(2n,n)

Answer 86

right, but thats only if you do the long division< right?

Answer 87

if you know to assume, 2x = y ?

Answer 88

or wait, you know that (0,0) is a solution and (1,1) is a solution, right>

Answer 89

yes...not assume but see if its correct by replacing 2x by y OR replacing y by 2x

Answer 90

(1,1) is known after we get other factors using x-2y

Answer 91

hmmm, yes but i mean is there something that would make us want to to do x = 2y

Answer 92

oh ,

Answer 93

tat is LUCK!

Answer 94

( 1,-1) is another solution then

Answer 95

OR if u knew the problem beforehand OR if u have wolframalpha open on ur browser.
No...-1 is not a POSITIVE integer :P

Answer 96

4x^2 + 9y^2 + 12xy -2x -15y +5 =0

How do you solve this btw?

Answer 97

from discriminant criteria for various conics see if its parabola/ellipse/hyperbola..

Answer 98

k. for positive integer solutions...ALL 2-VARIABLE EQUATION DO NOT HAVE positive integer solutions.....such questions are given after testing them..