Prob: Show that for any positive integer $a$, Diophantine equation $$x^3+x+a^2=y^2$$ has at least one solution $(x, y)$, where $x, y$ are positive integers.

Source: My teacher.

Attempt: First I tried $a=1$ and found the minimal solution $(x, y)=(72, 611)$, not a friendly one.

Now rewrite the equation as $$x(x^2+1)=(y-a)(y+a).$$ We hope that $x=b_{1}b_{2}, x^2+1=c_{1}c_{2}$ and $b_{2}c_{2}-b_{1}c_{1}=2a$. But how to determine these numbers? I got stuck here. One way promising is to relate to a Pell-type equation.

Another thought is to set $x$ to be some polynomial like $2t^2$ so that the left side can be factorized furthermore. Still little progress.

Please help.

  • $\begingroup$ The curve has the integer point $(0,a)$. At first glance this looks like an inflection point, but it's not. The tangent line intersects the curve at another point: $(1/4a^2,(1+8a^4)/8a^3)$. Perhaps these points can be used to generate another integer point? $\endgroup$
    – B. Goddard
    Aug 24, 2018 at 12:06
  • $\begingroup$ @B.Goddard Good point, but I don't think it can be very useful. We may consider it from the perspective of number theory rather than algebra. $\endgroup$
    – Juggler
    Aug 24, 2018 at 12:49

1 Answer 1


Your first idea is a great idea! When working this out, I had slightly different variables, but it is equivalent to your starting.

Suppose positive integers $b, c, d, e$ satisfy $y+a=bc$, $y-a=de$, $x=bd$, $x^2+1=ce$, $bc-de=2a$, $ce-(bd)^2 = 1$.

I propose the following lemma:

The sequence $k_n$ is defined as follows: $k \in \mathbb{N}$, $k_0 = 0$, $k_1=1$, $k_{n+2} = kk_{n+1} + k_{n}$ for $n \geq 0$. Then $k_{n-1} k_{n+1} - k_{n}^2 = (-1)^{n} \quad \forall n \geq 1$.


For $n=1$, this holds true. Now, assume true for $n$. Then for $n+1$,

$$\begin{aligned} k_{n} k_{n+2} - k_{n+1}^2 &= k_{n} (kk_{n+1} + k_{n}) - k_{n+1}^2\\ &= -k_{n+1} (k_{n+1} - kk_{n}) + k_{n}^2\\ &= -(k_{n-1} k_{n+1} - k_{n}^2)\\ &= -(-1)^n\\ &= (-1)^{n+1} \end{aligned}$$

Thus, proven.

Now, if we let $c=k_3 = k^2+1$, $e=k_5=k^4+3k^2+1$, $bd=k_4 = k(k^2+2)$, then from the above lemma, we have $ce-(bd)^2=1$. Letting $d=2a, k=4a^2$, we get $b=2a(k^2+2)$ and $bc-de=2a(k^2+2)(k^2+1)-2a(k^4+3k^2+1) = 2a$. Thus, we have the result

$$x=bd=8a^2(8a^4+1)$$ $$y=4a(8a^4+1)(16a^4+1)-a$$

as a construction for all $a$, and does satisfy the solution you provided for $a=1$.

  • $\begingroup$ Very through answer indeed! $\endgroup$
    – NoChance
    Jun 20, 2019 at 10:48

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