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

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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

Mathematics
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any thoughts/ideas?
couldn't u find a longer eq? :P
its a monstrous beast...

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Other answers:

:P
I think it is smth like (ax-by)^3 bu i can't tell wht r a and b
but*
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
(x-2*y)*(2*x-1+3*y)*(2*x-5+3*y)
(x-2y)(2x+3y-5)(2x+3y-1)=0
i'm not planning to deal with this..u do it...:P
i got it first, please medal
u rock cantorset! but we cant use comps!
how do you know i use computer
:)
@contorset: lol, its obvious. :P
hey how did u do tht?
well witout at computer its LUCK..
I didn't do it on my own, wolframalpha
maple
and how do u do it in comp?
haha... if u had a good luck u wud put x=2y and inspect!!!
you can do it in wolframalpha,
then u get (x-2y) and get others by dividing
i should not give out my secrets though
k. but tat doesnt solve the problem..
x=2y ? you mean substitute
@saubhik: You need a VERY good luck for that!
yes...
but see tat doesnt..solve the problem
we have to find positive integer solutions..
woww
but you can pick anything, x = 4y for instance
and i thought u guys were smart :P:P:P
i dont see any logic in this problem
ya cantorset that's truly possible but as i said we need luck and brain-breaking insights...
see..now how to find the solution?
for the first one(x-2y), x = 2I y=I
u can have infinite pairs (2n,n) for a positive integer n since x-2y is a factor...what for the other two?
yes
what for 2x+3y-5=0
its obvious (1,1) but are there others..THIS IS DIOPHANTINE!!
OK lets see, x + 3y/2 =5/2 x=(5-3y)/2 5-3y should be even,
so y is odd.
what about 2x+3y=1? yay...theres no positive integers here in this equation..so this is infertile:P
Why do you think so?
how can u get postive integers (x,y) from 2x+3y<5
there are none...:P
infertile..mother has no babies:P
that can be solved with euclids algorithm
Oh sorry, i dint read that 'positive' part! :P
this mother 2x+3y=5 has only one baby i.e. (1,1)
So our answer: (1,1) and (2n,n)
OMG!!! This is giving me headache only by reading ur explanations imagine if i had to solve this....:@@@@@@@@
So its done, (n,2n) U (1,1)
Openstudy rocks!! thnx evrybody!
Oops (2n,n). yes right
Going to deal with literature bye and have a good time :)
, amog, you wrote x= 1, y = 21?
@angela: buh-bye. @cantorset: It was an 'I' (eye)
for the first one(x-2y), x = 2I y=I
Yes i corrected it, didn't I?
hi cantorset...do u mind givin me an idea how to have a factor to the eqn..
suppose i have the baddest luck and go on inspecting from x=3y;x=4y....
wait, ... start over . why would i think to use x = 3y or anything like that
also its 3 linear, not a conic
or worser x=3y+7 etc..
yeah
you mean because the solution is ( ax + by ) ( dx + cy + ...
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
now clearly theres no constant term in this equation...so a linear factor (which must be present) must pass through the origin...
and so we start checking by x=y;x=2y;x=3y...
if the question does not mean to torment us we can get the factor in a few trials...
k..thnx evrybody for helping..:)
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?
what do you mean by constant term in the equation
term with no variables attched..
oh right
the final solution is 3 linears, right?
yups...family of 3 straight line
cool, so ok , theres no constant term, and clearly, (0,0) works
yups..try looking for x=y,2y,3y,...here u need luck
so that tells us that x = ay or y = ax must be one of the solutions?
yes..true
but only for a=2
i.e.(2n,n)
right, but thats only if you do the long division< right?
if you know to assume, 2x = y ?
or wait, you know that (0,0) is a solution and (1,1) is a solution, right>
yes...not assume but see if its correct by replacing 2x by y OR replacing y by 2x
(1,1) is known after we get other factors using x-2y
hmmm, yes but i mean is there something that would make us want to to do x = 2y
oh ,
tat is LUCK!
( 1,-1) is another solution then
OR if u knew the problem beforehand OR if u have wolframalpha open on ur browser. No...-1 is not a POSITIVE integer :P
4x^2 + 9y^2 + 12xy -2x -15y +5 =0 How do you solve this btw?
from discriminant criteria for various conics see if its parabola/ellipse/hyperbola..
k. for positive integer solutions...ALL 2-VARIABLE EQUATION DO NOT HAVE positive integer solutions.....such questions are given after testing them..
OK!
amogh u indian?
So here its a parabola
tats what wolframalpha says...
I didn't use wolframalpha for it? Afterall we in India are not allowed to use any calc! :P
And I like it!
in which class r u?
Just went into XII
wer r u frm?
kolkata passed 12
what are the possible shapes of a third degree x,y equation, a conic is second degree,
Cool, did you give the JEE?
may be: 1.family of straight lines.. 2. a strt line and a conic
gave it! but didnt get it:(
hope u get:)
Thanks :) So what plans?
very cool saubhik, didnt know that
dont know ....awaiting for AIEEE and WBJEE and ISI results...
whats AIEEE?
All India Engineering Entrance Examination (in india)
thnx..cantorset...
All the best buddy!
saubhik, i got a problem for ya
mayb ei should post new one
yups...
post new one..

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