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        riddles >> medium >> Perfect squares
        
(Message started by: fatball on Jan 22nd, 2006, 8:51pm)
Title: Perfect squares
Post by fatball on Jan 22nd, 2006, 8:51pm
Find all pairs of positive integers, x, y, such that x2 + 3y and y2 + 3x are both perfect squares.
Title: Re: Perfect squares
Post by Eigenray on Jan 23rd, 2006, 3:23am
Nice problem.  I guess I won't be getting any sleep before the semester starts this morning.
[hideb]Say x < y.  Now, for some r>0, we have
y2 + 3x = (y+r)2 = y2 + 2ry + r2,
or 3x=2ry+r2, and since x<y we must have r=1, 3x=2y+1.  So we may write
m2 = x2 + 3y = x2 + 3(3x-1)/2,
16m2 = 16x2 + 72x - 24 = (4x+9)2 - 105,
3*5*7 = 105 = (4x+9)2-16m2 = (4x+9+4m)(4x+9-4m),
so (4x+9-4m) = d, (4x+9+4m) = 105/d, for d=1,3,5, or 7 (taking m>0).  We can check these directly, or note that
d + 105/d = 2(4x+9) = 2 mod 8  implies d=1 mod 4,
but in any case the only possibilities are d=1, x=11, or d=5, x=1.  Recalling 3x=2y+1, this gives (x,y) = (11, 16) or (1, 1) as the only pairs with x<y.[/hideb]
Title: Re: Perfect squares
Post by Barukh on Jan 23rd, 2006, 4:38am

on 01/23/06 at 03:23:35, Eigenray wrote:
I guess I won't be getting any sleep before the semester starts this morning.

Does that mean we will be missing you frequently in the coming months?   :-[
Title: Re: Perfect squares
Post by Eigenray on Jan 24th, 2006, 6:08am
It means next time I stay up all night doing math, it'll probably be for a course.
Title: Re: Perfect squares
Post by fatball on Jan 24th, 2006, 10:31am
Eigenray, thanks for your effort. :-* Your answer is near perfect except that [hide]there is no need to assume x<y as they are symmetric equations.  That is also why you missed another pair of solution: (16,11)[/hide]

[hide]Here is another solution:
As x and y are positive integers, then we can write x2 + 3y and y2 +3x in the form (x + a)2 and (y + b)2 respectively where a, b are also positive integers.

Expanding and eliminating the squared terms, we get 2 linear simultaneous equations:
3y = 2ax + a2
3x = 2by + b2

Solving, we get
x = (2a2b + 3b2)/(9-4ab)
y = (2b2a + 3a2)/(9-4ab)

Since a and b are positive, the numerators in the above fractions are positive; for the denominators to be positive (in order to make x and y positive), we must therefore have ab = 1 or 2.

Possibilities are (a,b) = (1,1), (1,2), (2,1) => (x,y) = (1,1), (16,11), (11,16).[/hide]

Generalization: If there is no restriction on the sign of x and y (still integers), is there any additional solution?
Title: Re: Perfect squares
Post by JocK on Jan 24th, 2006, 10:58am

on 01/24/06 at 10:31:27, fatball wrote:
Eigenray, thanks for your effort. :-* Your answer is near perfect except that there is no need to assume x<y as they are symmetric equations.  That is also why you missed another pair of solution: (16,11)


Isn't it obvious that Eigenray had this in mind when he wrote "Say x < y" and "this gives (x,y) = (11, 16) or (1, 1) as the only pairs with x<y"..?
Title: Re: Perfect squares
Post by fatball on Jan 24th, 2006, 12:02pm
Well yes, his answer is perfect based on his assumption but it does not mean that it is a complete answer given the unneccessary assumptions...
Title: Re: Perfect squares
Post by JocK on Jan 24th, 2006, 12:08pm
Eigenray did not assume x < y. Rather, being aware of the permutation symmetry, he realised he only needed to investigate one of the two cases x<y and y<x.

 
Title: Re: Perfect squares
Post by fatball on Jan 24th, 2006, 12:16pm
Point taken.   :-*  :-*  :-*


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