Quadratic diophantine equations

Find two solutions to the following diophantine equation: $x^2-y^2=21$

In this case we have $n=21$. What it is necessary to do, then, is find the divisors of $21$. These are: $1, 3, 7$ and $21$.

Besides, the only way of multiplying them so that the result is exactly $21$ is:

  1. $1\cdot21=21$. The two are odd, therefore making $a = 1$ and $b = 21$ we get a solution by means of the formula $\displaystyle \begin{array}{c} x=\frac{a+b}{2} & y=\frac{a-b}{2}\end{array}$.

  2. $3\cdot7=21$ Here, both are also odd, therefore we will have another solution making $a = 3$ and $b = 7$ and substituting in the previous formula.

The two solutions are:

  1. If $a = 1$ and $b = 21$: $$x=\dfrac{1+21}{2}=11 \ \ \ \ y=\dfrac{1-21}{2}=-10$$ It is possible to easily check that this is a solution: $$11^2-(-10)^2=121-100=21$$

  2. If $a = 3$ and $b = 7$, then the solution is: $$x=\dfrac{3+7}{2}=5 \ \ \ \ y=\dfrac{3-7}{2}=-2$$ If we want, it is possible to check that it is really a solution: $$5^2-(-2)^2=25-4=21$$

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