# Proving this two equations are same and true

If $\sqrt{a} - \frac{1}{\sqrt{a}} = 1$, then $a + \frac{1}{a} = 3$.

Why this statement is true?

I tried to square the first equation, but it didn't work. I can't understand why there is a 3 in the second equation.

Assuming $a\geq0$, squaring both sides yields $$1=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)^2=(\sqrt{a})^2-2\sqrt{a}\frac{1}{\sqrt{a}}+\left(\frac{1}{\sqrt{a}}\right)^2=a-2+\frac{1}{a},$$ and hence $$a+\frac{1}{a}=1+2=3.$$
$(\sqrt{a}-\frac{1}{\sqrt{a}})^2 = 1^2$
$a-2+\frac{1}{a}=1$
$a+\frac{1}{a} = 3$.