Polynomial Diophantine Equation 2

Proposition : If the diophantine equation

$x^4+y^4 = z^2$

has integer solutions, then $xyz=0$

Proof : Assume that the above equation has solutions in positive integers. Let $(x,y,z)$ be a solution. We may assume that the $x,y$ and $z$ are relatively co-prime. Note that exactly one of $x$ and $y$ is even and $z$ is odd. W.L.O.G. let $x$ be even.

$(x^2)^2 + (y^2)^2 = z^2$

By Euclid's formula for Pythagorean Triples , we have,

$\displaystyle \left. \begin{array}{c}\ x^2=2ab\\ y^2=a^2-b^2\\ z=a^2+b^2 \end{array}\right\}$

where $\gcd(a,b) = 1$

Note that exactly one of $a$ and $b$ is even, since $x^2 = 2ab$ . Further, note that $b$ is even, since if $b$ is odd, then $a$ is even, $\implies y^2 \equiv a^2-b^2 \equiv -1 \pmod{4}$ , contradiction.

Now, $b^2+y^2 = a^2$

$\displaystyle \implies \left. \begin{array}{c}\ b=2cd\\ y=c^2-d^2\\ a=c^2+d^2 \end{array} \right\}$

where $\gcd (c,d) = 1$

$\implies x^2 = 2ab = 4cd(c^2+d^2)$

Since $c$ , $d$ and $c^2+d^2$ are mutually co-prime, we have,

$\displaystyle \left. \begin{array}{c}\ c=p^2\\ d=q^2\\ c^2+d^2=r^2 \end{array}\right\}$

$\displaystyle \implies p^4 + q^4 = r^2$

Note that the solution $(p,q,r)$ is a smaller solution than $(x,y,z)$ . Similarly, we can show that there are even smaller solutions. This sets up an infinite descent and hence $xyz=0 \ \square$

Now,

$\displaystyle x^8 + 6x^4y^2+y^4 = 2z^2$

$\displaystyle \implies 2(x^8 + 6x^4 y^2 + y^4) = (2z)^2$

$\displaystyle \implies (x^2+y)^4 + (x^2-y)^4 = (2z)^2$

By the Proposition, we have,

$\displaystyle (x^2+y)^4 \cdot (x^2-y)^4 \cdot (2z)^2 = 0$

$\displaystyle \implies y = x^2 \ ; \ z= 2x^4$

Thus the unique family of solutions to the diophantine equation is $(k,k^2,2k^4)$

For the given range, $\displaystyle k_{\max} = 7$ , $x = 7$ , $y = 49$ and $z = 4802$

$\displaystyle \therefore \max(x+y+z) = \boxed{4858}$

The polynomial above can be presented as: $(x^4+y^2)^2-z^2=z^2-4x^4y^2$ Then, $(x^4+y^2-z)(x^4+y^2+z)=(z-2x^2y)(z+2x^2y)$ The expression will be maximized for first bracket on the LHS equaling the first bracket on RHS. With a reodering we get $x^2=y ,2x^4=z$ now z has to be less or equal to 5000 for maximization. After a little bit of calculation we come to a conclusion that $x=7$ Substituting that back we get 4858.

Family of solutions : $(x,y,z)=(n,n^2,2n^4)$

n is any integer.

However I am not able to prove that it is unique.Any help is appreciated.

@Pi Han Goh @Mark Hennings