Check if there are nonzero natural numbers $n,x,y$ such that:
$$\sqrt{4n+5}+\sqrt{5n+1}+\sqrt{9n+4}= \frac{nx}{y}. $$Thank you in advance!
My ideas
So we can simply show that $4n+5,5n+1,9n+4$ are perfect squares. But I don't know what to do forward, how to show how each equal. Hope one of you can help me!
Best Answer
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.