An innocent-looking quadratic diophantine equation

Question

Can anyone give me a parametrization of the integer solution triples $(m, b, c)$ to the equation $$m^2 = 16bc + 8b + 8c+1 \quad \text{ ?}$$

I tried many reformulations of this problem. One of those uses the Principal Axes Theorem of Linear Algebra, i.e. by the substitution $$\left[ \begin{array}{r} d\\[5pt] e\\[5pt] f\\ \end{array} \right] \quad := \quad \frac{1}{\sqrt{2}} \left[ \begin{array}{rrr} \sqrt{2} & 0 & 0 \\[5pt] 0 & 1 & 1 \\[5pt] 0 & -1 & 1 \\ \end{array}\right] \left[ \begin{array}{c} m\\[5pt] b\\[5pt] c\\ \end{array} \right] $$ I arrived at the equation $$d^2-8e^2+8f^2-8\sqrt{2}\:e=1$$ whose sole merit is that is hasn't got any mixed terms. I then investigated the requirements for $e, f$ so that the back-substitution would deliver integral values for $b$ and $c$ (there is no problem with $m$ as $m = d$). However, I suspect this is a dead end.

Any suggestions are greatly appreciated.

Answer 1

I'm writing your problem as $y^2 - zx = -3$ In this version, a good starting solution is $(2,1,2).$

Another direction, and likely the only parameterization available, starts by taking integers $$ ad - bc = \pm 1 $$

There are ways to generate the modular group https://en.wikipedia.org/wiki/Modular_group#Finding_elements and determinant $-1$ can be found by negating the lower row, say.

Then create the automorphism matrix

$$ M = \left( \begin{array}{ccc} a^2 & 2ac & c^2 \\ ab & ad+bc & cd \\ b^2 & 2bd & d^2 \\ \end{array} \right) $$

To start with a solution $$ V= \left( \begin{array}{r} x \\ y \\ z \\ \end{array} \right) $$ multiply $MV$ to get a new solution. Note that, with $x,z$ even and $y$ odd, the product $MV$ is the same way.

I was a little worried about finding all solutions, test triples $(14,19,26)$ and variant $(2,19,182).$ Taking $a=-1, b=-13,c=1,d=12.$

for

$$ \left( \begin{array}{ccc} 1 & -2 & 1 \\ 13 & -25 & 12 \\ 169 & -312 & 144 \\ \end{array} \right) \left( \begin{array}{r} 14 \\ 19 \\ 26 \\ \end{array} \right) = \left( \begin{array}{r} 2 \\ 19 \\ 182 \\ \end{array} \right) $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Answer 2

Ternary indefinite quadratic form automorphism group and $f(x,y,z) = N$ for integer $N.$

If I have only a subgroup of the full group, if I can still connect every solution to some favorite ground solution then I know the entire solution set is parametrized by the group times the base. For more difficult forms there is the simple method of finding a few reflections, Cassels page 19, called symmetries.

I'm writing your problem as $y^2 - zx = -3$ In this version, a good starting solution is $(2,1,2).$

Another direction, and likely the only parameterization available, starts by taking integers $$ ad - bc = \pm 1 $$

$$ M = \left( \begin{array}{ccc} a^2 & 2ac & c^2 \\ ab & ad+bc & cd \\ b^2 & 2bd & d^2 \\ \end{array} \right) $$

$$ V_0= \left( \begin{array}{r} 2 \\ 1 \\ 2 \\ \end{array} \right) $$

for new solution $$ X = \left( \begin{array}{c} 2(a^2 + ac + c^2) \\ 2ab + ad+bc + 2cd \\ 2(b^2 + 2bd + d^2) \\ \end{array} \right) $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Now, we can show that every solution $(x,y,z)$ to $y^2 - zx = -3$ of positive integers with $y$ odd and $x,z$ even can be reduced to my favorite solution, $(2,1,2)$

First case, what if $z = x?$ As $y^2 - x^2 = -3 = (y-x)(y+x),$ also $y+x > |y-x|$ and $y+x = 3, y-x = -1.$ Thus $y=1, x=z=2.$ This is the ground solution I choose.

Second case, if $x > z > 0$ we just switch to $(z,y,x)$ and rename, the new $x$ is smaller then the new $z.$ This is accomplished with parameters $ a=d = 0, b=c=1$ and

$$ M = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array} \right) $$

Third case, the descent itself. Here $0 < x < z.$ We need to confirm that $y$ is in between.... well, $x,z$ are both $2 \pmod 4,$ so that $z \geq x+4,$ also $x \geq 2.$ So $$zx \geq (x+4) x = x^2 + 4x = x^2 + 2x + 2x \geq x^2 + 2x + 4.$$ Then $y^2 + 3 = zx,$ so $$ y^2 + 3 \geq x^2 + 2x + 4, $$ $$ y^2 \geq x^2 + 2x + 1 = (x+1)^2 $$ and $y > x$ So far we have $$ 0 < x < y < z $$ This time we take parameters $a=1, b = -1, c=0, d=1$ for $$ M = \left( \begin{array}{ccc} 0 & 0 & 0 \\ -1 & 1 & 0 \\ 1 & -2 & 1 \\ \end{array} \right) $$ so that the new triple is $$ (x, y-x, x-2y+ z) $$ Thus the new value of $y$ is strictly smaller, $x$ stays the same. Also we still have $y^2 + 3 = zx,$ so that the new $z$ is positive and strictly smaller then the old $z.$

As $y$ strictly decreases, we can do these steps only a finite number of times, new triple either $(z,y,x)$ or $ (x, y-x, x-2y+ z), $ until we reach a solution where neither $x < z$ nor $x > z$ holds, that is the root solution $(2,1,2).$ Hooray!

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

The advantage of having a group and matirx multiplication: given an initial column vector solution with large entries, carefully write out the string of matrices that take that initial $X$ to my $V_0.$ The reult of keeping track of the matrix product is some $W$ for which $WX = V_0.$ Then take $M = W^{-1}$ so that $X = M V_0$ Furthermore, the $a,b,c,d$ that give $W$ can just be inverted to give $M,$ the formula for $M$ far above is a homomorphism from the group of 2 by 2 (integer) matrices of determinant $\pm 1$ to the group of 3 by 3 matrices of determinant $\pm 1.$

Oh, earlier I deliberately took $b = -1$ in the matrix. It is likely that this step will be used many times in a row, so we might as well combine them:

$a=1, c=0, d=1$ and negative $b,$ $$ M = \left( \begin{array}{ccc} 0 & 0 & 0 \\ b & 1 & 0 \\ b^2 & 2b & 1 \\ \end{array} \right) $$ so that the new triple is $$ (x, y+bx, b^2 x + 2by+ z). $$ We choose $b < 0$ so that $y + bx > 0$ but $y + (b-1)x \leq 0.$ Thus the new $y$ has the minimum possible size. Also, the new $z$ is small, now $x > z > 0.$ So we do a switch step. In case you are familiar, this process is very close to Gauss reduction for positive binary quadratic forms. Also quite similar to the Euclidean algorithm for finding greatest common divisor of two numbers