Solve the equation C(n+2,4)=n^2-1 .

We'll write the formula for
combinations:

C(n,k) =
n!/k!(n-k)!

According to this formula, we'll write the
combinations for the given equation:

C(n+2,4) =
(n+2)!/4!(n+2-4)!

C(n+2,4) =
(n+2)!/4!(n-2)!

But (n+2)! =
(n-2)!*(n-1)*n*(n+1)*(n+2)

We notice in this product the
presence of the difference of squares:

(n-1)*(n+1) = n^2 -
1

C(n+2,4) = (n-2)!*(n^2 -
1)*n*(n+2)/4!*(n-2)!

We'll re-write the
equation:

(n^2 - 1)*n*(n+2)/4! = n^2 -
1

We'll divide by n^2 -
1:

n*(n+2)/4! = 1

n(n+2) =
4!

We'll remove the brackets and we'll substitute
4!:

n^2 + 2n = 1*2*3*4

n^2 +
2n - 24 = 0

We'll apply quadratic
formula:

n1 =
[-2+sqrt(4+96)]/2

n1 =
(-2+10)/2

n1 = 4

n2 =
-12/2

n2 =
-6

Since n has to be a natural number, the
only solution of the equation is n = 4.