One year ago, a father was 8 times as old as his son. Now, his age is the square of his son’s age. Find the age of the father

Let's denote the current age of the son as x (in years).

According to the problem, one year ago, the son was x - 1 years old and the father was 8 times as old, so the father was 8(x - 1) years old.

Now, the father's age is the square of his son's age, so the father is now x^2 years old.

But we also know that over one year, the father's age has increased by 1 year. So, 8(x - 1) + 1 = x^2.

Let's solve this equation:

8x - 8 + 1 = x^2, x^2 - 8x + 7 = 0.

This is a quadratic equation and can be factored into (x - 7)(x - 1) = 0.

Setting each factor equal to zero gives the possible solutions x = 7 and x = 1.

Since the son's age cannot be 1 (because one year ago he would have been 0 and the father would have been 0 as well, which contradicts the statement that the father was 8 times as old as his son), we discard x = 1.

So, the son is 7 years old.

Substituting x = 7 into the equation for the father's age gives the father's age as 7^2 = 49 years old.