Assume that all three radicands are perfect squares.
The least common multiple of 4, 5, and 9 is 180 so let's do arithmetic modulo 180. Though brute force, I found that all perfect squares must be in the set $\{0, 1, 4, 9, 16, 25, 36, 40, 45, 49, 61, 64, 76, 81, 85, 100, 109, 121, 124, 136, 144, 145, 160, 169\}$ (mod 180).
So $4n+5$ to be a perfect square, we must have $n \in \{1, 5, 10, 11, 14, 19, 20, 26, 29, 35, 41, 44, 46, 50, 55, 56, 59, 64, 65, 71, 74, 80, 86, 89, 91, 95, 100, 101, 104, 109, 110, 116, 119, 125, 131, 134, 136, 140, 145, 146, 149, 154, 155, 161, 164, 170, 176, 179\}$ (mod 180).
For $5n + 1$ to be a perfect square, we must have $n \in \{0, 3, 7, 12, 15, 16, 24, 27, 36, 39, 43, 48, 51, 52, 60, 63, 72, 75, 79, 84, 87, 88, 96, 99, 108, 111, 115, 120, 123, 124, 132, 135, 144, 147, 151, 156, 159, 160, 168, 171\}$ (mod 180).
For $9n+4$ to be a perfect square, we must have $n \in \{0, 4, 5, 8, 9, 13, 20, 24, 25, 28, 29, 33, 40, 44, 45, 48, 49, 53, 60, 64, 65, 68, 69, 73, 80, 84, 85, 88, 89, 93, 100, 104, 105, 108, 109, 113, 120, 124, 125, 128, 129, 133, 140, 144, 145, 148, 149, 153, 160, 164, 165, 168, 169, 173 \}$ (mod 180).
For all three to be perfect squares, you need $n$ to be in all three of these sets simultaneously. However, it turns out that the intersection of the sets is empty, so no such $n$ can exist.
Best Answer
It's clear that the numbers in the radical must be perfect squares,you could choose the largest number of $a,b,c$, WLOG, assume $a \ge b \ge c$.
Notice that $$a^2 \lt a^2+b+c+1 \le (a+1)^2$$ you only need to find when the equality holds.