Finding all pairs $(a,b)$ of positive integers such that $a^2+nab+b^2$ is a perfect square.

Question

When $n=2$, the question is trivial. Is there a general method to find all such pairs for $n\ge{3}$ and $n\in{\mathbb{N}}$?

Answer 1

Let $a^2+nab+b^2 = c^2$

Let $x = \dfrac ac$ and $y=\dfrac bc$.

Then we need to find rational soloutions $(x,y)$ to $x^2 + nxy + y^2 = 1$

We start with the particular solution $(x,y) = (-1,0)$ and "draw" the line $y = \dfrac uv(x+1)$, where $u,v$ are integers, through that point. It should intersect the hyperbola $x^2 + nxy + y^2 = 1$ at a rational point.

\begin{align} x^2 + nxy + y^2 &= 1 \\ x^2 + \dfrac uvnx(x+1) + \dfrac{u^2}{v^2}(x+1)^2 &= 1 \\ x &= \dfrac{v^2-u^2}{u^2+uvn+v^2} &\text{(We discarded $x=-1$.)} \\ y &= u\dfrac{nu+2v}{u^2+nuv+v^2} \end{align}

We get $a^2+nab+b^2 = c^2$ where

\begin{align} a &= v^2-u^2 \\ b &= nu^2+2uv \\ c &= u^2+nuv+v^2 \end{align}

If $(a,b)$ is a solution, then so too is $(-a,-b)$, $(b,a)$, and $(-b,-a)$. So all solutions can be characterized as

$$\{a,b\} \in \{v^2-u^2, nu^2+2uv\}, \quad c = u^2+nuv+v^2$$

Answer 2

Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.

Write the formula can someone come in handy. the equation:

$$Y^2+aXY+X^2=Z^2$$

Has a solution:

$$X=as^2-2ps$$

$$Y=p^2-s^2$$

$$Z=p^2-aps+s^2$$

more:

$$X=(4a+3a^2)s^2-2(2+a)ps-p^2$$

$$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$$

$$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$$

more:

$$X=(a+4)p^2-2ps$$

$$Y=3p^2-4ps+s^2$$

$$Z=(2a+5)p^2-(a+4)ps+s^2$$

more:

$$X=8s^2-4ps$$

$$Y=p^2-(4-2a)ps+a(a-4)s^2$$

$$Z=-p^2+4ps+(a^2-8)s^2$$

In the equation: $$X^2+aXY+bY^2=Z^2$$ there is always a solution and one of them is quite simple.

$$X=s^2-bp^2$$

$$Y=ap^2+2ps$$

$$Z=bp^2+aps+s^2$$

$p,s$ - integers asked us.

Answer 3

Another way to approach this is to use Hilbert's theorem 90.

As in steven gregory's reply aim to solve $$x^2+nxy+y^2=1$$ in the rationals. This is equivalent to $$N\left(x+\frac{ny}2+\frac y2\sqrt{n^2-4}\right)=1$$ where $N$ is the norm map from $\Bbb Q(\sqrt{n^2-4})$ to $\Bbb Q$. By Hilbert 90, the norm $1$ elements of the quadratic field are $$\frac{u+v\sqrt{n^2-4}}{u-v\sqrt{n^2-4}} =\frac{(u+v\sqrt{n^2-4})^2}{u^2-(n^2-4)v^2}$$ for rational $u$, $v$ not both zero. We can now grind out general formulae for $x$ and $y$, which will be equivalent to steven gregory's