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        riddles >> easy >> Pairs of perfect squares
        
(Message started by: NickH on Nov 21st, 2004, 9:33am)
Title: Pairs of perfect squares
Post by NickH on Nov 21st, 2004, 9:33am
Find all pairs of positive integers, x, y, such that

x2 + 3y
and
y2 + 3x

are both perfect squares.

Repeat for integers x, y, with no restriction on their sign.
Title: Re: Pairs of perfect squares
Post by Aryabhatta on Nov 21st, 2004, 3:48pm
I found the the postive integer version to be medium! Haven't yet tried the 'also consider negatives' version.

...
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Assume y [ge] x.
Cleary, if x = 1, then y2+3 is a perfect square only if y = 1. So (1,1) is the only solution with x = 1.

So assume x > 1.

x2+3y [ge] x2+3x > x2 + 2x + 1 as x > 1.

Thus we have that x2+3y  is at least (x+2)2 and therefore 3y [ge] 4x + 4.

This implies 3x [le] 3(3y-4)/4 < 4y.

Thus y2 + 3x < y2 + 4y < (y+2)2

Thus we must have that 3x = 2y+1.

Thus x2+3y = x2+ 9x/2 - 3/2

Since x2+3y is a perfect square we must have that it is either (x+1)2 or (x+2)2 (as (x+3)2 > x2+ 9x/2 - 3/2)

This gives x=1 or x=11.

Thus the only solutions are (1,1) and (11,16).
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...


If x and y can be negative... don't know... have to think about it.
Title: Re: Pairs of perfect squares
Post by Sir Col on Nov 21st, 2004, 4:57pm
I used a very similar approach...

::[hide]
First we note that x=y=1 is a solution.

W.L.O.G. assume that y>=x.

Clearly y2+3x < y2+4y < y2+4y+4 = (y+2)2.

As y2 < y2+3x <= (y+1)2, we deduce that y2+3x = (y+1)2, which leads to 3x = 2y+1.
Rearranging this we get, 3y = 9x-3/2.

Therefore, x2+3y = x2+9x/2-3/2 < (x+3)2.

Discounting the case already found where x=y=1, x2+3y >= x2+3x > x2+2x+1 = (x+1)2.

Therefore x2+3y = (x+2)2, giving 3y = 4x+4.

Solving 3y=4x+4 and 3x=2y+1, we get x=11, y=16.

Hence there are two distinct positive solutions.

Not including the infinite set of trivial cases where, x=0 and y=k2/3, I suspect, but haven't been able to prove yet, that there are six solutions inlvolving negatives:
x=-11, y=-7
x=-8, y=-5
x=-7, y=5
x=-5, y=8
x=-4, y=-4
x=-3, y=-3
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