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.