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?

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- Mar 5th 2013, 01:01 PMLockdownSolving 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? - Mar 5th 2013, 05:50 PMBobSacamanoRe: Solving diophantine equations
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.

- Mar 6th 2013, 06:28 AMLockdownRe: Solving diophantine equations
True, but I forgot to mention that I'm supposed to solve them for x,y,z with x,y,z>1.

- Mar 6th 2013, 06:49 AMjamixRe: Solving diophantine equations
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. - Mar 6th 2013, 07:09 AMLockdownRe: Solving diophantine equations
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! - Apr 30th 2014, 09:57 AMindividRe: Solving diophantine equations
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.