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Posted by **daphne** on Monday, January 10, 2011 at 10:39pm.

Any help or explanations would be wonderful!

- math -
**Reiny**, Monday, January 10, 2011 at 10:56pmthe 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

how about:

x+y=70

x-y=2

add them: 2x = 72

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.

how about (12+2)(12-2) ?

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

- math -
**Reiny**, Monday, January 10, 2011 at 11:00pmI left out

140 = 7x20

but x+y=20

x-y=7 has no integer solution, since 2x would have to be even.

- math -
**daphne**, Monday, January 10, 2011 at 11:05pmThank you very much!

- math -
**MathMate**, Monday, January 10, 2011 at 11:10pmSince

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 -
**MathMate**, Monday, January 10, 2011 at 11:14pmThanks Reiny, I left out combinations of cases where one or both of x and y is negative.

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