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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?

Josh
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2 Answers2

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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(nP));
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;
Mummy the turkey
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  • The small solutions $(11,3,122),(47,28,2593)$ are not shown. How is the list of solutions sorted? – Tomita Aug 15 '21 at 23:28
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    "...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. – jjagmath Aug 16 '21 at 01:12
  • @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. – Will Jagy Aug 16 '21 at 02:41
  • 1 1 2 \
    1 2 7 \
    11 3 122 \
    13 475 390794 \
    47 28 2593 \
    7199 4026 58941127 \\     – Will Jagy Aug 16 '21 at 03:22
  • @Will, I got $[47, 28, 2593]=[1316/28, 784/28, 2032912/28^2], [-11, 3, 122]=[ -33/3, 9/3, 1098/9 ].$ Thanks. – Tomita Aug 16 '21 at 07:05
  • @Tomita, the list was me simply asking computer algebra for the preimages of the $iP + jQ$ for small values of $i, j$ where $P, Q$ are the generators of the Mordell-Weil group. My restriction on the solutions is implicitly that there is no $p$ such that $v_p(x)= v_p(y) >0$ and $v_p(z) \geq 2$. I haven't checked, but it is also possible that (1) it hasn't been hit yet or (2) I missed some base points (the original curve should be smooth though, so I don't see why this would happen off hand) – Mummy the turkey Aug 16 '21 at 09:00
  • @jjagmath I don't see your point. The parametrisation of pythagorean triplets is exactly an isomorphism from $\mathbb{P}^1$ to a genus $0$ curve $C$ (you can dehomogenise it to get a one variable family if you wish). – Mummy the turkey Aug 16 '21 at 09:02
  • @jjagmath Of course, what you could be saying is that if there are only finitely many solutions then there is an obvious way of imposing conditions on lots of variables. But the whole point of this observation is that you never get something like the "well known parametrisation for pythagorean triplets". N.B., a (rational, sorry) parametrisation is defined to be a birational map from a projective space to a variety, so it is clear there is no parametrisation in this case. – Mummy the turkey Aug 16 '21 at 09:07
  • @Tomita I see now that the problem is basically that I was overzealous asking my computer to clear denominators, as noticed by Will Jagy. – Mummy the turkey Aug 16 '21 at 09:12
  • @WillJagy maybe this is not clear enough in my answer, but the point is that there are infinitely many points which avoid those trivialities (this is why I look at a curve in $\mathbb{P}(1,1,2)$ (it identified such points together), I've just listed a few of them. – Mummy the turkey Aug 16 '21 at 09:25
  • Mummy, are you saying that there are infinitely many integer nontrivial points? – Will Jagy Aug 16 '21 at 16:09
  • @Will indeed. If you get a rational one, just clear denominators by multiplying through by $u, u, u^2$ in $x,y,z$ where $u$ is the LCM of the denominators. This is a special case of the fact that rational points in a projective space are the same as the integral ones. The construction yields infinitely many rational points not related in this trivial way. – Mummy the turkey Aug 16 '21 at 17:18
  • Mummy, could you please try to find, with your method, a primitive integer triple not in my list? I agree that any rational point my be beefed up to be integral; I am not at all clear that this would lead to many distinct integer points with $\gcd(x,y,z) = 1.$ Let me do my list again... – Will Jagy Aug 16 '21 at 17:24
  • 1 0 1 \
    1 1 2 \
    1 2 7 \
    11 3 122 \
    13 475 390794 \
    47 28 2593 \
    7199 4026 58941127 \\ Looks better this way. I allowed zero or equal elements, I just demanded that gcd of all three be 1,     – Will Jagy Aug 16 '21 at 17:29
  • @Will How about $[ -32100744059569082693820895834319, 10810288733545862604250164736687, -1050149339544058566603186395792506486029740930747628808497108002 ]$ (or the less extreme example $[ -246121, 74983, -61353342962 ])$ – Mummy the turkey Aug 16 '21 at 17:41
  • Mummy, gp-pari says that one does not work. Please compare: ? a = 32100744059569082693820895834319 %1 = 32100744059569082693820895834319 ? b=10810288733545862604250164736687 %2 = 10810288733545862604250164736687 ? ? ? w = a^2 + 3*b^2 %3 = 1381044796685845581317668991586118065576644734757069315229401668 ? gcd(a,b) %4 = 1 ? factor(w) %5 = [ 2 2]

    [ 37 1]

    [ 236329 1]

    [ 7914229 1]

    [ 277750381 1]

    [ 39340145339533 1]

    [456593576393096675480082937 1]

     – Will Jagy Aug 16 '21 at 17:48
  • Let us continue this discussion in chat. – Mummy the turkey Aug 16 '21 at 17:50
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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:

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

Sam
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