2011/066) find near issoceles triangles

that is the 2 legs ar of length x, x+ 1
let legs be x and x+1 and hypotenuse y

x^2+ (x+1)^2 = y^2
or 2x^2 + 2x + 1 = y^2
or doubling

(4x^2+4x+2) = 2y^2
or (2x+1)^2 = 2y^2 -1
putting 2x +1 = m
we get 2y^2-m ^2= 1

now by trial we see that y = 5 and m = 7 is a solution
if (a,b) is a solution then (2a^2-b^2) = 1

so (a sqrt(2) – b) (a sqrt(2) + b) = 1
so | (a sqrt(2) – b) | = 1
cubing both sides ( sqaring shall not help as we get ( a+b- 2ab(sqrt(2)) this arises as |1| = |-1| = 1
(2a^3 + 3ab^2) sqrt(2) –(12a^2 + b^3) (the sqrt(2) term is positive and the non sqrt(2) is – ve)
So it is a solution
Starting with (5,7) we get (985,1393) or the triple is ( 696,697,845)
So we can get the terms as far as possible.
But It shall not give all solutions