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Thread: Solving diophantine equations

  1. #1
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    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?
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  2. #2
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    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.
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    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.
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  4. #4
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    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.
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    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.
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    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.
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