In this answer, I give an elementary solution of the Diophantine equation $$y^2=3x^4+3x^2+1.$$

In this post, the related equation $$y^2=3x^4-3x^2+1 \tag{$\star$}$$ is solved (in two different answers!) using elliptic curve theory. Is there an elementary method of proving that $x=1$ is the only positive rational solution to ($\star$)? In particular, I’d love to see a method using descent.

  • 2
    $\begingroup$ $x=-1$ is another solution. Note that the equation (*) is equivalent to $y^2 + (x^2-1)^3 = x^6$. Do we have a parametrization of the rational solutions to $y^2+z^3=x^6$? (If so then a similar method might apply to the first equation as well.) $\endgroup$ – Greg Martin Jan 29 at 1:42
  • $\begingroup$ Sorry — meant positive rational! Will edit. $\endgroup$ – Kieren MacMillan Jan 29 at 1:43
  • $\begingroup$ @Servaes: Not exactly — the complete answers on that post all use non-elementary methods. Fortunately, the answer below is elementary, so it answers my question perfectly! $\endgroup$ – Kieren MacMillan Feb 6 at 17:39

An elementary (and general) method

This is a little-known method which will solve many equations of this type. For this example it will show that the only positive rational solution is $p=1$. We start by letting $p=\frac{y}{x}$, where $x$ and $y$ are coprime integers. Thus we seek positive integer solutions of $$ x^4-3x^2y^2+3y^4=z^2.$$ This is one of a family of equations which can be dealt with together. (Apologies for using the letters for variables which I am used to when solving these types of equation.)


The only positive integer solutions of any equation of either of the forms $$ Ax^4+6x^2y^2+Cy^4=z^2,AC=-3\tag{1}$$ $$ ax^4-3x^2y^2+cy^4=z^2,ac=3\tag{2}$$ have $x=y=1$.


First note that, for either equation, we can suppose that $x,y,z$ are pairwise coprime since a common factor of any pair of $x,y,z$ would be a factor of all and cancellation can occur.

Also note that precisely one of $A$ and $C$ is divisible by $3$. Without loss of generality we can suppose that $3$ is a factor of $C$ and that $A\in \{\pm 1\}.$ Then $z^2\equiv A\pmod 3$ and so $A=1$. Similarly, we can suppose $a=1$.

An equation of form (1)

$x^4+6x^2y^2-3y^4=z^2$ can be rewritten, using completing the square, as $$\left (\frac{x^2+3y^2-z}{2}\right )\left (\frac{x^2+3y^2+z}{2}\right)=3y^4.$$ Since the two bracketed factors, $L$ and $M$ say, differ by the integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $z$ and $y$, a contradiction.

Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=3$ and $y=uv$, with $u$ and $v$ coprime. Then $$au^4+cv^4=x^2+3y^2=x^2+3u^2v^2.$$ Therefore $au^4-3u^2v^2+cv^4=x^2$, $ac=3$, an equation of form (2).

It is important to note that the mapping $(x,y,z)\rightarrow (u,v,w)$ is invertible. Only one solution set can map to $(u,v,w)$ by this process.

An equation of form (2)

Let $u,v,x$ be a pairwise coprime solution of $ u^4-3u^2v^2+3v^4=x^2$ and let $t$ be the greatest common divisor of $v$ and $2$. Then $(U,V,W)=(\frac{2u}{t},\frac{v}{t},\frac{4x}{t^2})$ is a pairwise coprime solution of $U^4-12U^2V^2+48V^4=W^2$.

This can be rewritten, using completing the square, as $$\left (\frac{U^2-6V^2-W}{2}\right )\left (\frac{U^2-6V^2+W}{2}\right)=-3V^4.$$ The bracketed factors, $L$ and $M$, are again coprime integers.Therefore $\{L,M\}=\{aX^4,cY^4\}$, where $ac=-3$ and $V=XY$, with $X$ and $Y$ coprime. Then $$aX^4+6X^2Y^2+cY^4=U^2,ac=-3,$$ an equation of form (1). Again, this mapping of solutions is invertible.

Fermat's infinite descent

We have seen that any positive integer solution $(x,y,z)$ of an equation of form (1) leads to another positive integer solution $(X,Y,Z)$, where $$Y=\frac{y}{tuX}.$$ We have to agree with Fermat that "there cannot be a series of numbers [positive integers] smaller than any given [positive] integer we please" and therefore the above process must lead to solutions with $tuX=1$.

It is now straightforward to plug $t=u=X=1$ back into the equations to see that there is only a simple loop consisting of the solutions $$x^4+6x^2y^2-3y^4=z^2,x=y=1,z=2$$ $$ x^4-3x^2y^2+3y^4=z^2,x=y=z=1.$$

  • 2
    $\begingroup$ Wonderful! As a bonus, I believe this answer — or at least the method — almost immediately solves my question math.stackexchange.com/questions/1680937/… (from the more general math.stackexchange.com/questions/1670607/…). Thanks! $\endgroup$ – Kieren MacMillan Jan 29 at 14:31
  • 2
    $\begingroup$ Yes you're right. Actually I think it's a lot more general than that and should deal with $ x^4\pm 3Nx^2y^2+3N^2y^4=z^2$ for $N$ square free. I'll check properly and perhaps post an answer to 1680937. I've not checked the details but I think these equations will have only trivial solutions unless $N$ is divisible by a prime congruent to $1$ modulo $3$. $\endgroup$ – S. Dolan Jan 29 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.