Infinite positive integer solutions of the equation: $x^2+x+1=(y^2+y+1)(n^2+n+1)$

Question

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.

Answer 1

HINT:

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.

Answer 2

To solve the Diophantine equation.

$$x^2+x+1=(z^2+z+1)(y^2+y+1)$$

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.

$$x=\mp((z+1)p^2+(z^2+z+1)(zs-p)s)$$

$$y=\mp((2z+1)p-(z^2+z+1)s)s$$

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