# math

posted by on .

Find all the ordered pairs of integers such that x^2 - y^2 = 140

Any help or explanations would be wonderful!

• math - ,

the hyperbola x^2 - y^2 = 140
can be factored to
(x+y)(x-y) = 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
x-y = 5
add them: 2x = 33 ---> x not an integer

x+y=70
x-y=2
x = 36, then y = 34
then (36+34)(36-34) = 140

(36,34), (36,-34), (-36,34) , (-36,-34)

a very limited number of cases.

so (12,2), (-12,2), (12,-2) and (-12,-2)

• math - ,

I left out
140 = 7x20
but x+y=20
x-y=7 has no integer solution, since 2x would have to be even.

• math - ,

Thank you very much!

• math - ,

Since
x²-y²
=(x+y)(x-y)
we are looking for two integers (x+y) and (x-y) 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 (x-y) must also be integers. This implies that (x+y) and (x-y) 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 (x-y) are either both even or both odd.
We have to address two cases:
If (x-y) is even, then
x+y
=(x-y)+2y
=2k+2y
=2(k+y)... so x+y is also even.
If (x-y) is odd, then
x+y
=(x-y)+2y
=(2k+1)+2y
=2(k+y)+1....so x+y is also odd.
QED

• math-correction - ,

Thanks Reiny, I left out combinations of cases where one or both of x and y is negative.