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I would like to find the integer solutions to the equation $x^4 + 3y^4 = z^2$.

I know this has infinitely solutions, because I noticed if $x=y$, we have $4x^4 = z^2 \Rightarrow 2x^2 = z$ (for positive integers). So for instance $(1,1,2), (2,2, 8), ...$ are solutions.

How to find all the integer solutions to this equation?

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    $\begingroup$ You might find the expression for all rational roots of $u^2+3v^2=1.$ Then engineer all examples where the numerators of $u,v$ are squares with common denominators. $\endgroup$ Aug 15, 2021 at 20:56
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    $\begingroup$ ...which can be found here, for example. $\endgroup$ Aug 15, 2021 at 20:56

2 Answers 2

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There are infinitely many such points- the proof below shows there is no closed from, but does show how to construct them. Let $$f(x,y,z) = x^4 + 3y^4 - z^2$$ and note that the curve, $C$, given by $f(x,y,z) = 0 $ in $\mathbb{P}(1,1,2)$ is a curve of genus $1$, and has the point $(1:1:2)$ (which you note) - this immediately shows there is no parametrisation, since there cannot be a map from $\mathbb{P}^1$ to ${C}$ which is non-constant.


In response to comments below I will clarify the usefulness of this construction.

Say that a solution to $f(x,y,z) = 0$ is primitive if there exists no prime $p$ such that $p$ divides $x$ and $y$, and $p^2$ divides $z$. Then primitive solutions to $f(x,y,z) = 0$ are in bijection with rational points on $C$.

Why? One direction is clear, a (primitive) integral solution gives rise to a rational point on $C$. If $(x,y,z)$ and $(u,v,w)$ are distinct primitive solutions then they give rise to distinct rational points since there exists no $a \in \mathbb{Q}$ such that $(x,y,z)=(au,av,a^2w)$ (this is the definition of the ambient weighted projective space).

Conversely, a consider a rational point $P = (x:y:z)$ on $C$. Then let $a$ be the LCM of the denominators of $x,y,z$ so we may assume that $x,y,z$ are integers (since $(x:y:z) \sim (ax:ay:a^2z)$ - the definition of weighted projective space). If $p$ divides $x,y$ and $p^2$ divides $z$ we note $(x:y:z) \sim (x/p:y/p:z/p^2)$ - so we obtain a primitive solution after finitely many operations.

Clearly distinct points (under the equivalence relation of weighted projective space) give rise to distinct primitive solutions.


In fact it is more convinient to take as our point $O$, the point $P = (1:0:1)$. Then setting $$x' = \frac{2x^2 + 2z}{y^2}$$ $$y' = 4\frac{x^3 + xz}{y^3}$$ gives an isomorphism between $C$ and the elliptic curve with Weierstrass equation $$(y')^2 = (x')^3 - 12(x')$$

but the point $(-2, 4)$ on $E$ has infinite order (this may be seen since, e.g., $3P = (-2/9, -44/27)$ has non-integral coordinates).

Indeed, after a $2$-descent on $E$ we see that the Mordell-Weil group has generators $(0 , 0), (-2, 4)$. The first has order $2$ and the second has infinite order as noted above.

Translating this back to $C$ gives infinitely many (primitive, integral) solutions $(x:y:z)$ by the discussion above.


Here is a list of the preimages of the points $n(0,0) + m(-2,4)$ for $-15 \leq n,m \leq 15$:

$$ [ 1, 0, 1 ], [ 1, 1, 2 ], [ 1, 2, 7 ], [ 11, 3, 122 ], [ 13, 475, 390794 ], [ 47, 28, 2593 ], [ 7199, 4026, 58941127 ], [ 246121, 74983, 61353342962 ], [ 3035713, 6824776, 81199358332033 ], [ 530296679, 498273129, 513813891524670482 ], [ 76359946657, 2413152950, 5830850177127262819399 ], [ 10962489040931, 11690411959381, 265471048509583706971158122 ], [ 1897729385992273, 3416601879194196, 20536759641325726509302231806753 ] $$

Of course, as you note, one can get (non-primitive) solutions for free by taking $ux$, $uy$ and $u^2z$ for any $u \in \mathbb{Z}$.

For interest's sake, here is some Magma code to verify:

P2w<x,y,z> := ProjectiveSpace(Rationals(), [1,1,2]);

C := Curve(P2w, x^4 + 3*y^4 - z^2);

E, phi1 := EllipticCurve(C, C![1,0,1]);
E2, phi2 := MinimalModel(E);

_, phi_inverse := IsInvertible(phi1*phi2);

P := Generators(E2)[2];
assert Order(P) eq 0;     //P has infinite order

n := 37;             // choose whatever you like here

cc := Coordinates(phi_inverse(n*P));
d := Denominator(cc[1]);
cc := [cc[1]*d, cc[2]*d, cc[3]*d^2];
cc := [Integers()!c : c in cc];       //Magma assumes rational
                                     //so we have to tell it it's integers.

GCD(cc);
cc;
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  • $\begingroup$ The small solutions $(11,3,122),(47,28,2593)$ are not shown. How is the list of solutions sorted? $\endgroup$
    – Tomita
    Aug 15, 2021 at 23:28
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    $\begingroup$ "...this immediately shows there is no parametrisation, since there cannot be a map from $\Bbb P^1$ to $C$". This doesn't show that there's no parametrization of the solutions. At most it proves that there isn't rational parametrization in one variable. But it could still be one with two or more variables and conditions imposed on them by some inequalities or modular restrictions (like the well known parametrization for the pythagorean triplets) as happens with many others diophantine equations. $\endgroup$
    – jjagmath
    Aug 16, 2021 at 1:12
  • $\begingroup$ @Tomita, with $x^4 + 3 y^4 = z^2,$ I see infinitely many solutions with $y=0$ or $y=x$ or $\gcd(x,y) >1$ However, I suspect there to be only finitely many triples that avoid the indicated trivialities. $\endgroup$
    – Will Jagy
    Aug 16, 2021 at 2:41
  • $\begingroup$ 1 1 2 \\\ 1 2 7 \\\ 11 3 122 \\\ 13 475 390794 \\\ 47 28 2593 \\\ 7199 4026 58941127 \\\ $\endgroup$
    – Will Jagy
    Aug 16, 2021 at 3:22
  • $\begingroup$ @Will, I got $[47, 28, 2593]=[1316/28, 784/28, 2032912/28^2], [-11, 3, 122]=[ -33/3, 9/3, 1098/9 ].$ Thanks. $\endgroup$
    – Tomita
    Aug 16, 2021 at 7:05
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The below equation has solution shown below:

$x^4 + 3y^4 = z^2$

$x=3m^2-n^2$

$y^2=4mn(3m^2+n^2)$

$z=(9m^4+18m^2n^2+n^4)$

RHS of $(y^2)$ is made a square at $(m.n)=(12,3)$ &

we get after removing common factors:

$(x,y,z)=(47,28,2593)$

Also for, $(m,n)=(1,1)$ we get:

$(x,y,z)=(1,2,7)$

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