Solving the Diophantine equation $k^2 = a^2 - 4b + 8$ for $k,a,b$

Question

We have the system of Diophantine equations in 4 variables

$$\begin{align} a & = p+q \\ b-2 &= pq \tag 1 \end{align}$$

We have

$$\begin{align} (p-q)^2 & = (p+q)^2 - 4pq \\ & = a^2 - 4(b-2) \tag 2 \end{align}$$

Putting $p-q = k$, we need to solve the 3-variable Diophantine equation

$$ k^2 = a^2 - 4b + 8 \tag 3 $$

Assuming we have a solution for Eqn. (3), we can then solve $$ p+q = a \\ p-q = k \tag 4 $$

I can get particular solutions for eg., if I set $b = 0$, then $k-a = \text{$d_i$, a divisor of 8}$ and $k+a = 8/d_i$.

How do I solve Eqn. (3) to obtain the general solutions?

If it helps to provide context, $a, b$ are the coefficients in a monic palindromic polynomial of degree 4.

$$ \begin{align} f(x) & = x^4 + ax^3 + bx^2 + ax + 1 \\ & = (x^2 + px + 1)(x^2 + qx + 1) \\ & = x^4 + (p+q)x^3 + (pq+2)x^2 + (p+q)x + 1 \end{align} \tag 5 $$

Equating the coefficients, we get the original Eqn. (1).

Note: WolframAlpha gives the following parametric solutions for Eqn (3):

$$ \begin{align} a & = 2 c_1, b = {c_1}^2 - {c_2}^2 + 2, k = 2 c_2, & \text{ where } c_1, c_2 \in \mathbb{Z} \\ a & = 2 c_1 + 1, b = {c_1}^2 + c_1 - {c_2}^2 + c_2 + 2, k = 2 c_2 - 1, & \text{ where } c_1, c_2 \in \mathbb{Z} \end{align} $$

Are these the only solutions?

Answer 1

If $a$ and $b$ are integer constants in your question, then $p$ and $q$ are obviously restricted and the pairs $(p,q)$ or $(q,p)$ are the roots of the quadratic equation $x^2-ax+(b-2)=0$ therefore $(p,q)$ is not always a pair of integers.

Thus, $a$ and $b$ cannot be considered as constants.

If $a,b,p,q$ are all integer variables, then you can construct the general solution set as follows:

$$\begin{align}&p=c_1\\ &q=c_2\\ &a=c_1+c_2\\ &b=c_1c_2+2\end{align}$$

where $c_1,c_2\in\mathbb Z$.

Thus, you can easily obtain the general solution you are looking for depending on the variables $a,b,k\,:$

$$ \begin{align}&a=c_1+c_2\\ &b=c_1c_2+2\\ &k=±(c_1-c_2)\end{align} $$ where $c_1,c_2\in\mathbb Z$.

Indeed, we see that:

$$ \begin{align}(c_1-c_2)^2=(c_1+c_2)^2-4(c_1c_2+2)+8\end{align} $$

Finally, observe that the solutions that Wolfram Alpha or other software generates for the equation $k^2=a^2-4b+8$ over integers, are a subset of the solution set we found.