# Find three different natural numbers that form an arithmetic progression and such that their product

**Find three different natural numbers that form an arithmetic progression and such that their product is a perfect square.**

1. Let a1, a2 and a3 be the first three terms of the arithmetic progression an:

a2 = a1 + d;

a3 = a1 + 2d, where

d – the difference of the progression.

2. By the condition of the problem, the product of numbers is a perfect square:

a1 * a2 * a3 = n ^ 2;

a1 (a1 + d) (a1 + 2d) = n ^ 2.

3. There are infinitely many solutions. Let’s select values for a1, a2 and a3:

1) 6k; 12k; 18k;

6k * 12k * 18k = 6 * (6k) ^ 3 = 6 ^ 4 * k ^ 3 = n ^ 2;

k must be square;

k = 1: 6; 12; eighteen;

k = 4: 24; 48; 72.

2) 12k; 18k 24k;

12k * 18k * 24k = 2 * 3 * 4 * (6k) ^ 3 = 6 ^ 4 * k ^ 3.

The same situation – k must be a square;

k = 1: 12; eighteen; 24;

k = 4: 48; 72; 96.

Answer: (6; 12; 18), (24; 48; 72), (12; 18; 24), (48; 72; 96), etc.