Two brothers write their ages, one after the other and get a four-digit number that is exactly the square of their father's age. Nine years later they rewrite their ages, in the same way, again obtaining a four-digit number that is the square of their father's age.

**What is the age difference between the two brothers?**

#### Solution

Sean **ab** Y **CD** the ages of the children and **n** the age of the father

According to the statement, the following two equations are fulfilled:

abcd = n²

abcd + 909 = (n + 9) ²

We solve the system as follows:

n² + 909 = n² + 9² + 2 · 9n

n² + 909 = n² + 81 + 18n

18n = 909 - 81

n = 828/18 = 46 years

abcd = 46² = 2116

The ages of the children are initially 21 and 16 and **the difference is 21-16 = 5 years**. Nine years later, the ages will be 30 and 25 that form the number 3025 which is the square of the father's age, 55 years.