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?
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.
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.