Quadratic Diophantine Equations - Solve by Completing the Square

We will tackle some Diophantine equations using the method of completing the square.

Consider the function $f(x) = ax^{2} + bx +c$. We are interested in knowing what values of $x$ make $f(x)$ a perfect square. We know that $f(x) = ax^{2} + bx +c$ can be written as $a\left(x+\frac{b}{2a}\right)^{2} + \frac{4ac-b^{2}}{4a}$.

For more information on how to transform $f(x)$ into this form, go here.

What values of $x$ make $f(x) = x^{2} + 6x + 8$ a perfect square?

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Solution 1: Let $f(x) = x^{2} + 6x + 8 = (x+3)^{2} - 1 = k^{2},$ where $x$ is an integer and $k$ is a non-negative integer. Observe that this can be rewritten as

$$\begin{aligned} (x+3)^{2} - k^{2} &= 1 \\ (x+3+k)(x+3-k) &= 1. \qquad (1) \end{aligned}$$

Then since both $x$ and $k$ are integers, there are only the following two possibilities for the left side of $(1)$ to be $1$ which is the right side:

$$\begin{aligned} x+3+k = 1, x+3-k &= 1 \\ \Rightarrow x &= -2, k = 0\\\\ x+3+k = -1, x+3-k &= -1 \\ \Rightarrow x &= -4, k = 0. \end{aligned}$$

That implies that the only integers $x$ such that $f(x)$ is a perfect square are $x = -2$ and $x = -4$. $_\square$

Solution 2: In this particular case, we see that: $(x+3)^{2} - k^{2} = 1$. The only squares that differ by $1$ are $1$ and $0$. So we look for those values of $x$ satisfying $(x + 3)^{2} = 1$, which gives $x = -2, x = -4.$ $_\square$