# “Descent” on binary quadratic forms?

Let's say I have the Diophantine equation $$x^2+3n^2 = y^2+3z^2. \tag{\star}$$ where $n$ is a known integer, and we're trying to determine solutions in integers $x,y,z \ge 1$. Rewrite ($\star$) as $$x^2 - 3z^2 = y^2 - 3n^2 = k, \tag{\dagger}$$ where $k$ is some unknown integer. For simplicity's sake, let's consider the case $n=1$, so we have $$x^2 - 3z^2 = y^2 - 3(1)^2 = k, \tag{\Diamond}$$ Evidently, $(y,1)$ must be the fundamental solution of the equation $U^2-3V^2=k$. Now either $\lvert z \rvert=1$ implying $\lvert x \rvert=\lvert y \rvert$, or $\lvert z \rvert>1$ implying $\lvert x \rvert > \lvert y \rvert$. Let's assume the more interesting latter case.

My question is: can Pell equation machinery — in particular, some form of descent/ascent — be used to solve the original equation, or at least give formulas for $x$ and $z$ in terms of $y$ and $1$ (or some other permutation of the elements)?

I realise we don't know how far “up the ladder” the solution $(x,z)$ is, but I thought there might be some known and effective results in this direction.

EDIT: Also of interest to me would be an answer in the special case where $y$ is an unknown integer with a known parameterization, e.g. $y=ab+cd$ for positive [but possibly unknown] integers $a,b,c,d$.

The complete solution to,

$$x_1^2+mx_2^2 = x_3^2+mx_4^2\tag{1}$$

is given by,

$$(a c + b d m)^2 + m(a d - b c)^2 = (a c - b d m)^2 + m(a d + b c)^2\tag{2}$$

known as the Brahmagupta-Fibonacci identity.

We wish to solve $ad-bc = \pm n$ in the integers for some constant $n$. We can do this as a Pell-like eqn by using the substitution $a,b,c = q^2,\, p+q,\, p-q$ to get,

$$ad-bc = p^2-(d+1)q^2 = \pm n\tag{3}$$

for some arbitrary $d$. For some constant $\pm n$, there is an indefinite number of $d$ that will do, each one of which will lead to a different Pell-like eqn with an infinite number of integer solutions. (Note that this Pell-like eqn doesn't depend on $m$ at all which, in your example, was $m=3$.) The $x_i$ can then be expressed in terms of $p,q$.

More simply, we can just solve $ad-bc = n$ for $d$, let $a=1$, then an explicit formula is,

$$(b^2 c m + b m n + c)^2 + m n^2 = (b^2 c k + b m n - c)^2 + m(2b c + n)^2\tag{4}$$

where one term ($x_2$) remains a constant n regardless of the three free variables $b,c,m$.

One particular solution.

I have already said, where the formula in General.

For the equation: $$3x^2+y^2=z^2+3n^2$$

If you use the solutions of the Pell equation. $p^2-3s^2=1$

And we have a number $y,n$ known. Moreover, any sign.

Then:

$$x=(p^2+3s^2)n-2psy$$

$$z=6psn-(p^2+3s^2)y$$

• Yes, this is a particular solution. Unfortunately, in my case, I don't know $y$, only $n$. However, I'll see if this formula could still be used to help with a descent argument. – Kieren MacMillan Sep 25 '14 at 12:09
• @KierenMacMillan $y,n$ - can be any. – individ Sep 25 '14 at 13:18

We can at least say that there are infinitely many solutions with the parametrization:

$$(x, y, z) = (7n, 2n, 4n)$$

which gives us

$$(7n)^2 + 3n^2 = (2n)^2 + 3(4n)^2 = 51n^2$$

Of course, I'm sure this does not cover all the possible solutions. For example, if $n$ is even, then another possible parametrization is:

$$(2n)^2 + 3n^2 = \left(\frac{n}{2}\right)^2 + 3\left(\frac{3n}{2}\right)^2 = 7n^2$$