Using owl’s last example (and a lot of trial and error in Excel) which gave the four square roots as 3000, 3003, 3006, 3009, I determined that the average of the squares of three roots minus 2/3 of the square of the other root gave me one of the four distinct positive numbers. This worked for all four combinations.

I wondered if this would work for any four consecutive multiples of 3.

So, for x, x+3, x+6, x=9 I arrived at

x²/3 – 6x – 39, x²/3 + 6, x²/3 + 6x + 33, x²/3 + 12x + 42 as my 4 distinct positive integers.

This only works for x >= 24 or I end up with a negative number for the first expression.

E.g. For 24, 27, 30, 33 I got 9, 198, 369, 522

This would seem to give an infinite number of solutions of four consecutive multiples of 3.

However, my grade 10 maths is not enough to know why this works (or how to come up with solutions without consecutive multiples of 3).