Surprise Diophantine

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PhilG

#1 h Oct 9, 2009, 9:05 AM • 1 Y

Y by Adventure10

Show that the following equations have an infinite number of simultaneous solutions.

$a^4 + b^4 + c^4 = 2d^4$ and $a^2 + b^2 + c^2 = 2d^2$ with $\gcd(a,b,c,d) = 1$ and $a,b,c,d \in \mathbb{N}$

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phymax

#2 h Oct 13, 2009, 7:48 AM • 1 Y

Y by Adventure10

$\frac{a^4+b^4+c^4}{2}=\left ( \frac{a^2+b^2+c^2}{2} \right )^2$

$a^4-2a^2(b^2+c^2)+(b^2+c^2)^2=0$
solve quadratic in $a$ .
$a^2=\frac {2(b^2+c^2)\pm \sqrt{2^2(b^2+c^2)^2-4(b^2+c^2)^2}}{2}=b^2+c^2$
QED

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PhilG

#3 h Oct 13, 2009, 9:41 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

There is a sign error in your second equation. It should be

$a^4 - 2a^2(b^2 + c^2) + (b^2 - c^2)^2 = 0$

so your solution is incomplete

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phymax

#4 h Oct 13, 2009, 11:26 AM • 2 Y

Y by Adventure10, Mango247

thanks for the correction..
$a^{2}=\frac{2(b^{2}+c^{2})\pm\sqrt{2^{2}(b^{2}+c^{2})^{2}-4(b^{2}-c^{2})^{2}}}{2}$
$=b^{2}+c^{2}\pm \sqrt{4b^2c^2}=b^2+c^2\pm 2bc=(b\pm c)^2$

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pco

#5 h Oct 13, 2009, 11:57 AM • 2 Y

Y by Adventure10, Mango247

phymax wrote:

thanks for the correction..
$a^{2} = \frac {2(b^{2} + c^{2})\pm\sqrt {2^{2}(b^{2} + c^{2})^{2} - 4(b^{2} - c^{2})^{2}}}{2}$

$= b^{2} + c^{2}\pm \sqrt {4b^2c^2} = b^2 + c^2\pm 2bc = (b\pm c)^2$

But this is not finished !

If $a^2=(b\pm c)^2$ , you get $\frac{a^4+b^4+c^4}2=\left(\frac{a^2+b^2+c^2}2\right)^2$ but you are not sure that $\frac{a^2+b^2+c^2}2$ is a perfect square ( $d^2$ )

for example : $b=2$ , $c=1$ and $a=b-c=1$ imply $\frac{a^2+b^2+c^2}2=3$ and $\frac{a^4+b^4+c^4}2=9$ but there is no integer $d$ such that $a^2+b^2+c^2=2d^2$

IMHO

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PhilG

#6 h Oct 13, 2009, 12:05 PM • 2 Y

Y by Adventure10, Mango247

So close, but there are solutions.

E.G $(a,b,c,d) = (3,5,8,7)$

You just have to show there are infinitely many.

So far you have shown that
$a \pm b \pm c = 0$ and $a^2 + b^2 + c^2 = 2d^2 \implies a^4 + b^4 + c^4 = 2d^4$

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bboypa

#7 h Oct 15, 2009, 3:29 AM • 2 Y

Y by Adventure10, Mango247

PhilG wrote:

Show that the following equations have an infinite number of simultaneous solutions.
$a^4 + b^4 + c^4 = 2d^4$ , $a^2 + b^2 + c^2 = 2d^2$ with $\text{gcd}(a,b,c,d) = 1$ and $(a,b,c,d) \in \mathbb{N}^4$

$\displaystyle \left(\sum_{cyc}{a^2}\right)^2 = 2\left(\sum_{cyc}{a^4}\right)$ so $\displaystyle \prod_{cyc}{(a + b - c)} = 0$ . So it is enough to set $c: = a + b$ and show that the equation $a^2 + ab + b^2 = d^2$ iff $\left(\frac {2a + b}{2d}\right)^2 + 3\left(\frac {b}{2d}\right)^2 = 1$ has infinitely many coprime solution in $\mathbb{Z}^3$ . But it is straghforward and well known that the elliptic curve $X^2 + 3Y^2 = 1$ has infinitely many points in $\mathbb{Q}^2$ (if more details are necessary look here for all class solutions).

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PhilG

#8 h Oct 15, 2009, 8:50 AM • 2 Y

Y by Adventure10, Mango247

Thank you for completing the problem

The result can also be seen as an application of the multiplicative property of quadratic forms. I.e.

$(x^2 + 3y^2)(z^2 + 3t^2) = (xz - 3yt)^2 + 3(xt + yz)^2$

We require a solution to $u^2 + 3v^2 = 4d^2$ .
To get one just take any odd integers $x,y$ and take $d$ from $x^2 + 3y^2 = 2d$
Using the multiplicative identity we find $u = x^2 - 3y^2, v = 2xy \implies u^2 + 3v^2 = 4d^2$

In fact we can repeat the multiplication of solutions to get higher powers for the quadratic form which can be used to show that there are an infinite number of solutions for $a^4 + b^4 + c^4 = 2d^{2n}$ for any $n \in \mathbb{N}$

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