Admittedly I only have a very brief introduction to diophantine equations under my belt, but for 2. it seems you can let (x,y,z) = (1,t,t) which will be a solution for any integer t.
Regarding 1).........there is a quick solution and a LONG solution.
Simply plug in some small integers and see if the relation holds. If your lucky and observant, you'll immediately see that (x,y) = (0,1) is a solution!
Rearranging, we have y^2 - 1 = (y-1)(y+1) = 4x^3
As the RS is even, we must have that both (y-1) AND (y+1) be even (do you know why its "AND" as apose to "OR"?).................Divide by 4 on both sides then let z = (y-1) / 2.
We have z(z+1) = x^3 (why?).
Now as gcd(z,z+1) = 1, it follows that BOTH z and z+1 must be a cube, but this implies that there are integers "a" and "b" such that (z+1) - z = a^3 - b^3 = 1.
The only solution to a^3 - b^3 = 1 is a = 1 and b = 0 (why?). Hence z = 0 and x = 0, which implies that y = 1.
Of course for the equation
There is a particular solution:
But interessuet is another solution:
If you use the solution of Pell's equation:
Making formula has the form:
number and any given us.
The most interesting thing here is that the numbers it Pythagorean triple.
This formula is remarkable in that it allows using the equation
Allows you to find Pythagorean triples with a given difference.
Pretty is not expected relationship between the solutions of Pell's equation and Pythagorean triples.