Find three consecutive natural numbers if you know that the square of the smaller one is 65

Find three consecutive natural numbers if you know that the square of the smaller one is 65 less than the product of the other two numbers.

Let’s denote by x the first number from the given sequence of three consecutive natural numbers.

Then the second number from this sequence will be equal to x + 1, and the third number from this sequence will be equal to x + 2.

According to the condition of the problem, the square of the smaller of these numbers is 65 less than the product of the other two numbers, therefore, we can make the following equation:

x ^ 2 + 65 = (x + 1) * (x + 2).

We solve the resulting equation:

x ^ 2 + 65 = x ^ 2 + x + 2x + 2;

x ^ 2 + 65 = x ^ 2 + 3x + 2;

3x = 65 – 2;

3x = 63;

x = 63/3;

x = 21.

Find two other numbers:

x + 1 = 21 + 1 = 22;

x + 2 = 21 + 2 = 23.

Answer: the required numbers are 21, 22 and 23.