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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.

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    $\begingroup$ $x=-1$ 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
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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.)

Theorem

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 $x=y=1$.

Proof

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 $$x^4+6x^2y^2-3y^4=z^2,x=y=1,z=2$$ $$ x^4-3x^2y^2+3y^4=z^2,x=y=z=1.$$

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    $\begingroup$ 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
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    $\begingroup$ 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

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