Well, I have a number $n$ that is given by:


I want to find $x\in\mathbb{Z}$ such that $n$ is a perfect square.

I found the following solutions:


Is there a way to prove that this a complete set of solutions? So I mean that the solutions given in formula $(2)$ are the only ones?

My work:

  • We know that: $$ 1 + 12x^2 \left(1+x \right) \ge 0 \space \Longleftrightarrow \space x \ge -\frac{1+2^{-2/3}+2^{2/3}}{3} \approx -1.07245 \tag3 $$ So we know that for $x<-1$ there are definitely no solutions.
  • $\begingroup$ maybe this helps, maybe not, but you can simply rewrite it in another form: $k(k+1) = l(l + 1) \cdot 3l$ $\endgroup$
    – Anatoliy R
    Dec 4, 2019 at 23:17
  • 2
    $\begingroup$ Dupe of this recent unanswered question. Is this question from a contest? $\endgroup$ Dec 5, 2019 at 1:18
  • $\begingroup$ @BillDubuque yes it is $\endgroup$ Dec 5, 2019 at 8:41
  • $\begingroup$ A link to the contest page, please. We need to ascertain that it is not from a live contest. Also, you can find the answer from a database of Elliptic curves over $\Bbb{Q}$. That is, if you know how to conver this into a Weierstrass form. $\endgroup$ Dec 5, 2019 at 12:44
  • $\begingroup$ @JyrkiLahtonen Can you help me finding the elliptic curve that is represented by my equation to solve this problem? $\endgroup$ Dec 5, 2019 at 12:57

2 Answers 2


$y^2=1+12x^2(1+x) \implies (12 y)^2 = (12 x)^3 + 12 (12 x)^2 + 144$

Magma code for positive $y$ only:

S:= IntegralPoints(EllipticCurve([0,12,0,0,144]));
for s in S do
  x:= s[1]/12;
  if x eq Floor(x) then
    print "(",x,", ",Abs(s[2]/12),")";
  end if;
end for;


( -1 ,  1 )
( 0 ,  1 )
( 1 ,  5 )
( 4 ,  31 )
( 6 ,  55 )
  • $\begingroup$ The question is, are those the only solutions? $\endgroup$ Dec 6, 2019 at 8:39
  • $\begingroup$ @Jan of course. For small coefficients Magma calculator found all solutions. $\endgroup$ Dec 6, 2019 at 8:44
  • $\begingroup$ is that proven? If yes where? $\endgroup$ Dec 6, 2019 at 8:51
  • $\begingroup$ Recording a link to the LMFDB entry of this curve. The equation there is the minimal Weierstrass model of this curve, so we need to figure out the transformation to get there, and to use the list of integer points given there. Anyway, their data is reliable. Unfortunately it is too late for me to work out the transformation. $\endgroup$ Dec 6, 2019 at 21:57


n is odd, let $n=(2m+1)^2$ ; m∈Z, then:

$4m(m+1)=12x^2(1+x)$$m(m+1)=3 x^2 (1+x)$

Suppose $m=3t$, then:



For example $n=31^2= (6\times 5 +1)^2$ or $n=55^2=(6\times 9+1)^2$

Therefore there may be more solutions. Now let $m+1=3t$ then:

$(3t-1)3t=3x^2(1+x)$$(3t-1) t=x^2(1+x)$$n=(2m+1)^2= (6t-1)^2$

For example $n=5^2=( 6\times 1 -1)^2$

In this case there may also be some solutions.

Moreover n can be the square of negative numbers, so it can have general forms $n=[±(6t+1)]^2$ or $n=[±(6t-1)]^2$.We have to show whether equation $x^3+x^2=3t^2±t$ can have infinite solutions or limited number of solutions in Z.


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