Author |
Topic: Pairs of perfect squares (Read 903 times) |
|
NickH
Senior Riddler
Gender: 
Posts: 341
|
 |
Pairs of perfect squares
« on: Nov 21st, 2004, 9:33am » |
 Quote  Modify
|
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.
|
|
 IP Logged |
Nick's Mathematical Puzzles
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: Pairs of perfect squares
« Reply #1 on: Nov 21st, 2004, 3:48pm » |
Quote Modify
|
I found the the postive integer version to be medium! Haven't yet tried the 'also consider negatives' version.
...
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).
...
If x and y can be negative... don't know... have to think about it.
|
| « Last Edit: Nov 21st, 2004, 3:50pm by Aryabhatta » |
IP Logged |
|
|
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825
|
|
Re: Pairs of perfect squares
« Reply #2 on: Nov 21st, 2004, 4:57pm » |
Quote Modify
|
I used a very similar approach...
::
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
::
|
|
IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print