The important thing to realize in order to solve this problem is that if square roots are multiplied, you can group them together. In math terms:
sqrt(a)*sqrt(b)=sqrt(ab)
So, for example, sqrt(32)*sqrt(2)=sqrt(32*2)=sqrt(64)=8.
In this equation, we have sqrt(n^2-1)*sqrt(n^2+1). Using the rule described above, we can combine them into one square root:
sqrt(n^2-1)*sqrt(n^2+1)=sqrt((n^2-1)(n^2+1))
The expression (n^2-1)(n^2+1) may not look like anything we know at first, but recall the difference of squares property:
(a-b)(a+b)=a^2-b^2
(n^2-1)(n^2+1) is actually a difference of squares, with a=n^2 and b=1. Since this simplifies into a^2-b^2, we can say:
(n^2-1)(n^2+1)=(n^2)^2-(1)^2=n^4-1
We've just simplified sqrt(n^2-1)*sqrt(n^2+1) into sqrt(n^4-1). Our problem now looks like:
sqrt(n^4-1)=sqrt(n^4-1)
Hm...both sides are the same! What does that mean? It means the equation has infinitely many solutions! Any value of n will satisfy this equation.
