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couldn't u find a longer eq? :P
its a monstrous beast...
I think it is smth like (ax-by)^3 bu i can't tell wht r a and b
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
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
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!
but see tat doesnt..solve the problem
we have to find positive integer solutions..
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?
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..
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..
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?
but only for a=2
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
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..
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...
All India Engineering Entrance Examination (in india)
All the best buddy!
saubhik, i got a problem for ya
mayb ei should post new one
post new one..