math
posted by daphne .
Find all the ordered pairs of integers such that x^2  y^2 = 140
Any help or explanations would be wonderful!

the hyperbola x^2  y^2 = 140
can be factored to
(x+y)(xy) = 140
140 = 2x5x2x7
or in pairs:
14x10 or 28x5 or 2x70, or 35x4
so we are looking for any two numbers x and y
so that their sum x their difference is 140, using only the numbers above
e.g x+y=28
xy = 5
add them: 2x = 33 > x not an integer
how about:
x+y=70
xy=2
add them: 2x = 72
x = 36, then y = 34
then (36+34)(3634) = 140
(36,34), (36,34), (36,34) , (36,34)
a very limited number of cases.
how about (12+2)(122) ?
so (12,2), (12,2), (12,2) and (12,2) 
I left out
140 = 7x20
but x+y=20
xy=7 has no integer solution, since 2x would have to be even. 
Thank you very much!

Since
x²y²
=(x+y)(xy)
we are looking for two integers (x+y) and (xy) that have a product of 140, and such that x and y are integers.
We can start by enumerating the factors of 140:
140*1
70*2
35*4
28*5
20*7
14*10
Since x and y are both integers, and (x+y) and (xy) must also be integers. This implies that (x+y) and (xy) must be either both odd or both even (for proof, see end of post).
The only pairs that meet this criterion are 14,10 and 70,2, which when solved give (x,y)=(12,2), or (36,34).
Given x, y ∈ℤ, proof that (x+y) and (xy) are either both even or both odd.
We have to address two cases:
If (xy) is even, then
x+y
=(xy)+2y
=2k+2y
=2(k+y)... so x+y is also even.
If (xy) is odd, then
x+y
=(xy)+2y
=(2k+1)+2y
=2(k+y)+1....so x+y is also odd.
QED 
Thanks Reiny, I left out combinations of cases where one or both of x and y is negative.