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.
