I'm sorry, I don't really know how to go about anyone's suggestions so continuing with Case 3 I thought since the last is p^2 - q^2 + 4(p+q) then from the equation p^2 - y^2 = 2 so the equation becomes 2 + 4(p + q) and therefore since all positive, all are greater than two unless p and q...
Bah, I forgot that natural numbers aren't negative. Would you include zero as a natural?
So from (x-y)(x+y) = 2,
Case 1: ( x = y )
Left side = (0)(x.x) = 0 ≠ 2
Case 2: ( y > x )
Left side is negative integer multiplied by positive integer which is equal to a negative integer which ≠...
So using (x-y)(x+y) = 2 I thought of:
if x and y are equal then left hand side is equal to zero (contradiction)
if y is greater than x
I thought that the left hand side would always be negative but this won't be the case as say
x = -3 and y = 1 , then left hand side is equal to 8 which...
I'm trying to prove that there are no natural numbers x and y that satisfy the equation x^2 - y^2 = 2.
I tried to solve it by contradiction and so I assume that x and y are rational numbers and both x and y can be written in the form (a/b) where it's in its simplest form and a and b are both...