Could anybody help me some ideas on the below problem:

Let $n$ be a positive integer. Prove that there are infinite pairs of positive integer $(x\, y)$ such that $$x^2+x+1=(y^2+y+1)(n^2+n+1).$$ Thanks in advance.

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    As per the box to the right of the big textbox you typed your question into, "Provide details. Share your research." – Eric Towers May 1 at 1:36

To solve the Diophantine equation.


It is necessary to use the solutions of the Pell equation. $p^2-(z^2+z+1)s^2=\pm1$

Then the solutions can be written as follows.



For positive you need to take decisions at $-1$.


Fix $n$. You get a quadratic equation in $x$, $y$. You should reduce it to a Pell equation as follows:

$$(4 x^2 + 4 x + 4 ) = (n^2 + n+1)(4 y^2 + 4 y + 1)\\ (2x+1)^2 + 3 = N \cdot ((2y+1)^2 + 3)\\ (2x+1)^2 - N (2y+1)^2 = 3(N-1)\\ a^2 - N b^2 = 3(N-1)$$

You have a solution for the last equation, $a=2n+1$, $b=1$, coming from the obvious equality $(n^2 + n+1)=(n^2 + n+1)(0^2 + 0+1)$.

Consider now a solution for the Pell equation $$A^2 - N B^2 =1$$ with $A$, $B$ positive integers ($N = n^2 + n+1$ is odd, not a square). Note that $A$, $B$ have opposite parities. Now $$(a'+b'\sqrt{N}) = (A+B\sqrt{N})(a+b\sqrt{N})$$ satisfies again $$a'^2 -N b'^2 = 3(N-1)$$ and $a'$, $b'$ are odd positive integers. Since there are infinitely many such $A$, $B$, we get infinitely many solutions $(a,b)$ odd positive integers.

  • Thanks. I also thought of this idea. However, do we have any other ideas without using Pell type equation? – Võ Quốc Bá Cẩn May 1 at 13:56
  • @Võ Quốc Bá Cẩn: Hm.., I am not sure. The equation is equivalent to the one $a^2 - N b^2 = 3(N-1)$, and any two solutions of these are connects by a solution of the Pell equation. So you somehow have to know that Pell has infinitely many, or equivalently,at least one nontrivial solution. So, if you had two solutions for your equation ( positive, then you have infinitely many. But that is about as hard as Pell. – orangeskid May 1 at 14:03

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