Recent content by lisamane

Earlier you said let y = x + a which if you put in the equation it should be x2 - (x + a)2 = 2 How did it come to (x + a)2 - x2 = 2 ?

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...