# Solving diophantine equations

• Mar 5th 2013, 12:01 PM
Lockdown
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?
• Mar 5th 2013, 04:50 PM
BobSacamano
Re: 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, 05:28 AM
Lockdown
Re: 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, 05:49 AM
jamix
Re: 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, 06:09 AM
Lockdown
Re: 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, 08:57 AM
individ
Re: Solving diophantine equations
Of course for the equation $\displaystyle X^2+Y^2=Z^2+t$

There is a particular solution:

$\displaystyle X=1\pm{b}$

$\displaystyle Y=\frac{(b^2-t\pm{2b})}{2}$

$\displaystyle Z=\frac{(b^2+2-t\pm{2b})}{2}$

But interessuet is another solution: $\displaystyle X^2+Y^2=Z^2+1$

If you use the solution of Pell's equation: $\displaystyle p^2-2s^2=\pm1$

Making formula has the form:

$\displaystyle X=2s(p+s)L+p^2+2ps+2s^2=aL+c$

$\displaystyle Y=(p^2+2ps)L+p^2+2ps+2s^2=bL+c$

$\displaystyle Z=(p^2+2ps+2s^2)L+p^2+4ps+2s^2=cL+q$

number $\displaystyle L$ and any given us.

The most interesting thing here is that the numbers $\displaystyle a,b,c$ it Pythagorean triple. $\displaystyle a^2+b^2=c^2$

This formula is remarkable in that it allows using the equation $\displaystyle p^2-2s^2=\pm{k}$

Allows you to find Pythagorean triples with a given difference.

$\displaystyle a=2s(p+s)$

$\displaystyle b=p(p+2s)$

$\displaystyle c=p^2+2ps+2s^2$

$\displaystyle b-a=\pm{k}$ Pretty is not expected relationship between the solutions of Pell's equation and Pythagorean triples.