Quadratic diophantine equations

The quadratic diophantine equations are equations of the type: $a x^{2} + b x y + c y^{2} = d$ where $a$ , $b$ , $c$ and $d$ are integers, and we ask the solutions $x$ and $y$ to be integers.

Nevertheless, we are only going to see quadratic diophantine equations of this kind here: $x^{2} - y^{2} = n$ with $n$ as any integer.

In this case, just as before,the equation may either have no solution or more than one solution. Nevertheless, the condition for this diophantine equation to have a solution is simpler: if $n$ can be written as the product of two numbers that both even, or both odd, then there will be solution. For example:

Example

If $n = 4$ we have $n = 2 \cdot 2$ , and both are even, therefore the equation $x^{2} - y^{2} = 4$ has a solution.

Example

If $n = 15$ , we have $n = 3 \cdot 5$ , $3$ and $5$ are odd both, therefore the equation $x^{2} - y^{2} = 15$ has solution.

Example

If $n = 6$ , and the divisors of $6$ are $1, 2, 3$ and $6$ .

Besides, for the result of multiplying them to be $6$ , we either have to use $1 \cdot 6$ , or $2 \cdot 3$ , since no other way of writing $6$ as product of $2$ (positive) integers is possible.

In neither case is it satisfied that both numbers are even or odd, so the equation $x^{2} - y^{2} = 6$ has no solution.

Let's suppose now that $n = a \cdot b$ , where $a$ and $b$ even or odd both. Then a solution is given by: $\begin{matrix} x = \frac{a + b}{2} & y = \frac{a - b}{2} \end{matrix}$

Example

For example, in the case that has been seen before $n = 4 = 2 \cdot 2$ , we see that $a = 2$ and $b = 2$ , therefore a solution is: $\begin{matrix} x = \frac{2 + 2}{2} = 2 & y = \frac{2 - 2}{2} = 0 \end{matrix}$

Let's observe that in the case there is a solution, it cannot be unique, since it is possible that $n$ admits another decomposition as the product of two even or odd numbers.

For example, if $n = 16$ , we know that $n = 2 \cdot 8$ (both are even) but also $n = 4 \cdot 4$ (both are even), and each of these two representations of $n$ gives a solution other than the diophantine equation.