Problems from Quadratic diophantine equations

Development:

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

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

Solution:

The two solutions are:

1) If $a=1$ and $b=21$: $$x={\displaystyle \frac{1+21}{2}}=11y={\displaystyle \frac{1-21}{2}}=-10$$ 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: $$x={\displaystyle \frac{3+7}{2}}=5y={\displaystyle \frac{3-7}{2}}=-2$$ 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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