Find the no. of ordered pairs (x,y) such that x^2+3y & y^2+3x botha are perfect squares.

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I think he means that all numbers involved are integers. If x= y, you can prove that the only solutions is x = y  =1.

You can even prove that x and y have got to be relatively prime.

take that back: i am not sure anymore that x and y have to be relatively prime

This is a nice problem. Only this is not the typical JEE kind. Not, that I discourage you guys from trying out such problems, but you are much better off doing Calculus and Coord Geometry than these.

I am solving for x and y both positive integers here.

Ok, we are given that both  are perfect squares.

Now, with x² of course being a square, adding 3y makes it a perfect square. That means 3y must be of the form  where m is a positive integer as  is a perfect square. Similarly we have the relation

Thus, we have the simulataneous equations

Eliminating y we get

m and n are positive. Since x>0, we must have 9>mn

Hence, the possibilities are

1. m = n =1

2.  m=2, n=1

3.m=1, n=2

For these we get the ordered pairs (1,11), (16,11) and (11,16) as the only solutions among positive integers.      

Correction: the first ordered pair is (1,1) and not (1,11)

again a wrong one !!

see , u wrote

" Now, with x² of course being a square, adding 3y makes it a perfect square. That means 3y must be of the form  where m is a positive integer ....."

consider x=3, m=5

so y=16 /3 makes it !!

now we have 2mx + m^2 = 2*5*3 +25 = 55 which is not 3y ( i.e. 16 )

So ur whole argument breaksdown !!!

You are proving yourself to be a priceless idiot.

Quite obviously, the difference of two squares need not always be divisible by 3. So not every m qualifies.

Honestly, abhished, I think you are way out of your depth here, so dont even attempt to poke holes.

The problem is done and the thread closed. Find some other pastime

Good Luck!

Plz  atleast try to read the problem before u start copy-pasting from another book .

The problem asks to find all ordered pairs (x,y) , who has told u that they are going to be integers( unless ur careless assumption ) ?There are other holes in the proof also !!

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there you go making accusations. For your info I spent the good part of the day trying various approaches of solving the problem and then arrived at this solution. In the past whenever I have filched a solution, I have named my source. A recent example: http://www.goiit.com/posts/list/algebra-sir-please-reply-me-70919.htm#350600 (Notice the courtesy I.A. Maron)

All ordered pairs? there are infinite with x = 0, y = 3ⁿ where  n is odd. and with m<0, n<0 there are so many solutions, so obvious

That is why I chose positive integers where it comes out neatly.

Now stop wasting everyone's time

 x=0 , y= 3^n are still integral solutions  (it is not quite general as  there are vaious n's that don't satisfy it n= 2 for example)

But that is not my point !!

Ok, for the sake of argument let us solve it for integral x,y s . But how on earth do u assume that they are ( 3y)  going to be ONLY of the form 2mx +m^2 .? ( assuming m positive )

It is certainly a sufficient condition but not a necessary one !!

I think u have certain intelligence to get my point .

So ur method can't gurantee that u have found all the integral solns.

Finally , I am still a larner and my education is not complete . But I think that u have firm conviction that u have completed all ur mathematics lessons and have nothing to learn anymor