# Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?

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.

• $x=-1$ 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.) – Greg Martin Jan 29 at 1:42
• Sorry — meant positive rational! Will edit. – Kieren MacMillan Jan 29 at 1:43
• @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! – 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.)

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

• 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! – Kieren MacMillan Jan 29 at 14:31
• 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 $3$. – S. Dolan Jan 29 at 14:46