# Does an elementary solution exist to $x^2+1=y^3$?

Prove that there are no positive integer solutions to $$x^2+1=y^3$$ This problem is easy if you apply Catalans conjecture and still doable talking about Gaussian integers and UFD's. However, can this problem be solved using pre-university mathematics?

I am talking about elementary number theoretical solutions. Do they exist?

• @Alizter If ab is a square and a and b are coprime, then both are squares. Oct 2, 2014 at 21:21
• Oct 2, 2014 at 23:20
• I think the answer is "YES" — see my answer/proof, below. Nov 9, 2014 at 20:25
• I just found a second, far more elegant, elementary solution. I think I'm going to write this one up and submit it for publication. Thanks for the inspiration! Feb 24, 2016 at 20:49
• @PluckyBird: I’m putting together the final version of my paper now. It includes elementary solutions — most new — for the equations $X^3=Y^2+k$ with $-4 \le k \le 4$. In several cases (including this one), I’m including multiple elementary proofs. Jun 29, 2020 at 21:35

Lemma. If $$a$$ and $$b$$ are integers such that $$3a^4+3a^2+1=b^2\!$$, then $$a=0$$ and $$b=\pm 1$$.

Proof. We may, without loss of generality, assume that $$a$$ and $$b$$ are non-negative. Assume, contrary to the lemma, that $$a \ne 0$$ and $$b > 1$$ are integers satisfying the equation. Evidently $$b$$ is odd, say $$b=2c+1$$ for an integer $$c \ge 1$$. Hence \begin{align*} 3a^4 + 3a^2+1 &= (2c+1)^2 \\ 3a^2(a^2+1) &= 4c(c+1). \end{align*} Since $$4 \nmid 3(a^2+1)$$, regardless of the parity of $$a$$, we must have $$2 \mid a^2$$; hence $$a$$ is even, say $$a=2d$$ for an integer $$d \ge 1$$. Now \begin{align*} 3d^2(4d^2+1) &= c(c+1). \end{align*} We now show that this equation has no solutions with $$c,d \ge 1$$. Assume to the contrary that there exist integers $$p,q,r,s \ge 1$$ such that \begin{align*} 3d^2 &= pq, &&& c &= pr, \\ 4d^2+1 &= rs, &&& c+1 &= qs. \end{align*} Since $$\gcd(p,q) \mid \gcd(pr,qs) = \gcd(c,c+1)=1$$ implies $$\gcd(p,q)=1$$, we may write $$d=uv$$ with $$\gcd(u,v)=1$$ and consider the two possible cases.

Case 1: $$p=u^2$$ and $$q=3v^2$$. Hence $$c=pr = u^2r$$, and by substitution $$c+1=u^2r+1 = 3v^2s$$. On the other hand, $$rs-1 = 4d^2=4u^2v^2$$, and adding these two relations yields \begin{align*} (rs-1)+(u^2r+1) &= 4u^2v^2 + 3v^2s \\ r(s+u^2) &= v^2(4u^2+3s). \end{align*} Since $$d=uv$$, the equation $$rs = 4d^2+1$$ forces $$\gcd(v,r)=\gcd(s,u)=1$$. Hence $$v^2 \mid (u^2+s)$$, and $$\gcd(s+u^2,4u^2+3s) = \gcd(s+u^2,3(s+u^2)+u^2)=\gcd(s+u^2,u^2) = \gcd(s,u^2)=1$$. Thus we conclude $$r=4u^2+3s$$ and $$v^2=s+u^2$$. Substituting now gives $$\begin{equation*} r = 4u^2+3s = 4u^2+3(v^2-u^2) = u^2+3v^2. \end{equation*}$$ Multiplying, $$4u^2v^2+1=rs=(u^2+3v^2)(v^2-u^2)$$, which implies $$u^4+1=3v^2(v^2-2u^2)$$. But $$3 \nmid u$$ (because $$3 \mid q$$ and $$\gcd(p,q)=1)$$, so $$u^4+1 \equiv 2\!\pmod{3}$$, a contradiction.

Case 2: $$p=3u^2$$ and $$q=v^2$$. Now $$c+1=pr+1=qs$$ gives $$3u^2r+1=v^2s$$. We have $$rs-1=4d^2=4u^2v^2$$, and adding the two relations yields \begin{align*} (rs-1) + (3u^2r+1) &= 4u^2v^2 + v^2s \\ r(s+3u^2) &= v^2(4u^2+s). \end{align*} As before, considering common factors leads to the conclusion $$s+3u^2=v^2$$ and $$\begin{equation*} r=4u^2+s =4u^2+(v^2-3u^2) = u^2+v^2. \end{equation*}$$ Multiplying, $$4u^2v^2+1 = rs = (u^2+v^2)(v^2-3u^2)$$, which is $$3u^2(u^2+2v^2)=(v^2-1)(v^2+1)$$. Since $$3$$ can never divide $$v^2+1$$, we deduce $$v^2-1=3u_1^2w_1$$ and $$v^2+1=u_2^2w_2$$ where $$u=u_1u_2$$ and $$u^2+2v^2=w_1w_2$$. As $$4 \nmid (v^2+1)$$, we deduce that $$u_2$$ is odd, $$w_2 \equiv 1\text{ or }2\!\pmod{4}$$, and $$\gcd(u_2,w_1)=1$$. Adding yields $$2v^2 = (v^2-1)+(v^2+1) = 3u_1^2w_1 + u_2^2w_2$$, and by substitution \begin{align*} w_1w_2 &= u^2+2v^2 \\ &= u_1^2u_2^2+ (3u_1^2w_1 + u_2^2w_2) \\ w_2(w_1-u_2^2) &= u_1^2(u_2^2+ 3w_1). \end{align*} Since $$\gcd(w_1-u_2^2,u_2^2+3w_1)$$ divides the sum $$4w_1$$, and $$\gcd(w_1,u_2)=1$$ and $$u_2$$ is odd, we deduce $$w_1-u_2^2=u_1^2$$. Now $$w_2 = u_2^2+3w_1 = u_2^2+3(u_2^2+u_1^2) = (2u_2)^2+3u_1^2 \equiv 0\text{ or }3\!\pmod{4}$$, contradicting $$w_2 \equiv 1\text{ or }2\!\pmod{4}$$ and proving the lemma.

Theorem. The equation $$X^2 +1 = Y^3$$ has only one integer solution, namely $$(x,y)=(0,1)$$.

Proof. Assume $$x$$ and $$y$$ are integers such that $$x^2+1 = y^3$$. Considering the equation modulo $$3$$, we quickly deduce that $$x^2 = (3z+1)^3−1 = 9z(3z^2+3z+1)$$ for some integer $$z$$. Since $$\gcd(z,3z^2+3z+1)=1$$, both $$z$$ and $$3z^2+3z+1$$ must be squares. Write $$z=a^2$$ for an integer $$a$$, and hence for some integer $$b$$ we have $$b^2 = 3z^2+3z+1 = 3a^4+3a^2+1$$. Now the lemma forces $$a=0$$, so that $$z=a^2=0$$, and $$y=3z+1=1$$, as claimed.

• +1 Nicely done. The main thing to prove the lemma was choosing factors $p$, $q$, $r$ and $s$. My approach to the equation $3a^4+3a^2+1=b^2$ was showing contradiction through infinite descent and Pell-Fermat equation but I was stuck for several days and couldn't get it. I'm glad that this question has finally been solved using elementary methods :) Nov 18, 2014 at 11:12
• @MathGod: I think a proof by descent would be more enlightening than mine, and would probably lead to a more general method applicable to more equations than this ad hoc method. We should keep looking for that! Nov 18, 2014 at 11:57
• There is a small gap, which will be easy to fill (and the fix might actually give us a hint towards a proof by descent) — I'll post an updated/corrected answer soon. Nov 21, 2014 at 15:18
• @awllower: $3u^2=v^2-s$, not $r^2-s$, right? Dec 23, 2019 at 4:21
• @MathGod: For a Pell+descent proof, see <jstor.org/stable/24496717>. Jun 28, 2020 at 0:41

An alternative proof and a generalisation

As stated in the accepted answer, it is sufficient to prove that $$3a^4+3a^2+1=b^2$$ has no positive integer solution. With a change of notation we shall prove the following generalisation by using the same method used for Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?

Theorem

No equation of either of the forms $$Ax^4-6x^2y^2+Cy^4=z^2,AC=-3\tag{1}$$ $$ax^4+3x^2y^2+cy^4=z^2,ac=3\tag{2}$$ has a positive integer solution.

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.

Conclusion

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 $$Y=\frac{y}{tuX}.$$ Any monotonically strictly decreasing series of positive integers must terminate and so the above process must lead to solutions with $$tuX=1$$.

It is now straightforward to plug $$u=X=1$$ and thus $$U=2$$ back into the equations to obtain the contradiction $$1-6Y^2-3Y^4=4.$$