Let x and y be integers x > y, such that x² - y² = 400
x² - y² = 400
(x - y)(x + y) = 400
(x - y) represents the smaller factor of 400 and (x + y) represents
the larger factor.
400 can be the product of two integers only in these ways:
(x - y)(x + y) = 400 = 1*400, where (x - y) = 1 and (x + y) = 400
(x - y)(x + y) = 400 = 2*200, where (x - y) = 2 and (x + y) = 200
(x - y)(x + y) = 400 = 4*100, where (x - y) = 4 and (x + y) = 100
(x - y)(x + y) = 400 = 5*80, where (x - y) = 5 and (x + y) = 80
(x - y)(x + y) = 400 = 8*50, where (x - y) = 8 and (x + y) = 50
(x - y)(x + y) = 400 = 10*40, where (x - y) = 10 and (x + y) = 40
(x - y)(x + y) = 400 = 16*25, where (x - y) = 16 and (x + y) = 25
(x - y)(x + y) = 400 = 20*20, where (x - y) = 20 and (x + y) = 20
We eliminate some of these:
If x + y is even, then so is x - y
If x + y is odd, then so is x - y
The product of two even numbers is even and
the product of two odd numbers is odd, so since
400 is even, we can rule out the three cases where
an odd number appears as either factor. We can also rule
out the case where both factors = 20, since that would
give the pair 20 and 0, but 0 is not a positive integer.
so we have only thse 4 possible cases:
1. (x - y)(x + y) = 400 = 2*200, where (x - y) = 2 and (x + y) = 200
2. (x - y)(x + y) = 400 = 4*100, where (x - y) = 4 and (x + y) = 100
3. (x - y)(x + y) = 400 = 8*50, where (x - y) = 8 and (x + y) = 50
4. (x - y)(x + y) = 400 = 10*40, where (x - y) = 10 and (x + y) = 40
In case 1, we have this system of equations:
with solution x = 101, y = 99
In case 2, we have this system of equations:
with solution x = 52, y = 48
In case 3, we have this system of equations:
with solution x = 29, y = 21
In case 4, we have this system of equations:
with solution x = 25, y = 15
Edwin
