Thread: Solving diophantine equations

Hello everyone,

I'm having some trouble with the two following diophantine equations:

1. solve y^2= 4x^3 +1 for x,y in Z
2. show that x^2+y^2 = z^2 +1 has infinitely many solutions

Can anyone help?

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.

True, but I forgot to mention that I'm supposed to solve them for x,y,z with x,y,z>1.

Regarding 1).........there is a quick solution and a LONG solution.

QUICK METHOD:

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!

LONG WAY..........

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.

Ah, I see! Thank you.

I had already found k(k+1) = x^3, but I didn't realize that that implies that both k and k+1 are cubes.
Thanks!