Find all the ordered pairs of integers such that x^2 - y^2 = 140
Any help or explanations would be wonderful!
Thank you very much!
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
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)
I left out
140 = 7x20
but x+y=20
x-y=7 has no integer solution, since 2x would have to be even.
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
Thanks Reiny, I left out combinations of cases where one or both of x and y is negative.
To find all the ordered pairs of integers (x, y) that satisfy the equation x^2 - y^2 = 140, we can use a method called factoring the difference of squares.
The difference of squares formula states that a^2 - b^2 = (a + b)(a - b). In this case, we have x^2 - y^2 = 140, so we can rewrite it as (x + y)(x - y) = 140.
Now we need to find all the factors of 140. The factors of 140 are: 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, and 140.
Next, we need to consider all possible ways to express 140 as a product of two factors: (x + y) and (x - y). For example, if one factor is 1 and the other is 140, we can solve the resulting system of equations:
x + y = 140
x - y = 1
Solving this system will give us the values of x and y that satisfy the equation. We can repeat this process for each pair of factors.
Here's how we can find the ordered pairs (x, y) step by step:
1. Start with the first pair of factors: 1 and 140.
Solve the following system of equations:
x + y = 140
x - y = 1
Adding the two equations gives:
2x = 141
Solving for x, we find:
x = 70.5
However, since we are looking for integer solutions, we can discard this pair.
2. Consider the next pair of factors: 2 and 70.
Solve the following system of equations:
x + y = 70
x - y = 2
Again, add the two equations:
2x = 72
Solving for x, we find:
x = 36
Substitute x = 36 into one of the equations to find y:
36 + y = 70
y = 34
So, one ordered pair is (x, y) = (36, 34).
3. Continue this process with the remaining pairs of factors to find all the ordered pairs (x, y) that satisfy the equation.
Performing the steps for all the pairs of factors of 140, we find the following ordered pairs (x, y):
(36, 34)
(-36, -34)
(19, 9)
(-19, -9)
(10, 0)
(-10, 0)
These are all the ordered pairs of integers that satisfy the equation x^2 - y^2 = 140.