Quadratic diophantine equations

The quadratic diophantine equations are equations of the type: $ax^2+bxy+cy^2=d$ 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:

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

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.

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: $$\displaystyle \begin{array}{c} x=\frac{a+b}{2} & y=\frac{a-b}{2}\end{array}$$

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: $$\displaystyle \begin{array} {c}x=\frac{2+2}{2}=2 & y=\frac{2-2}{2}=0\end{array}$$

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.