A Hidato, but with a complex twist!

Question

Today's Hidato puzzle has a pretty complex twist. What is the twist, you ask?

This is what the twist is:

There are three different colors:

• Yellow: Represents a prime number
• Green: Represents a square number ($\sqrt x$ must produce a number $\alpha$ with $\alpha\in\mathbb N$)
• Purple: Represents a number in the Fibonacci Sequence if that number is not already represented with another color.

Also, there are only 6 numbers to begin with.

Here is the puzzle:

The goal of Hidato is to fill the grid with a series of consecutive numbers adjacent to each other orthogonally or diagonally. All tiles are required to be filled in.

I have tested this out, and am able to confirm that this truly does have only one unique solution. (took me around 5 minutes)

Text version for colorblind users (best I could do, if someone could make this better then please do unless this is good enough):

-------------------------
|1  |Y  |   |Y  |12 |   |
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|   |G  |Y  |G  |Y  |   |
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|Y  |Y  |P  |Y  |G  |Y  |
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|   |   |G  |   |18 |   |
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|   |Y  |G36|35 |Y  |P  |
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|Y  |30 |   |   |P  |   |
-------------------------

Answer 1

Solution:

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Step by step:

1:

A key thing to note that isn't explicitly stated is that all primes are coloured, and same for the other colours. Using this, the purple numbers must be 8, 21, and 34. The 34 must be next to the 35.
So knowing this, 2 and 3 can be easily placed to start. Now if the bottom right purple is 8, it won't be able to reach 12, so it must be the other purple, with the bottom right purple being 21.

The two yellows touching the 12 must be 11 and 13, and if the top is 13 then 15 would have to be on green, but as it is not square, the top yellow is 11, which also places the 10. The two greens below are then a 4/9 pair. Top right must be 14 and 15 as neither of them are prime or square.

2:

16 can now be placed from 15, and this means the other green is 25. Now looking at the order of the 4 and 9, if the 4 is on the right then there is no place for a 7, so it must be on the left. The 6 must be the white cell, which gives one option for the 5, 6 and 7.

3:

The yellows next to 16 must be 17 and 19, and for 19 to reach 21 the 19 must be on the right. 22 and 23 can also be placed at the bottom, as well as 24 to get to 25. We know the two remaining yellows are 29 and 31, and for 30 to get to 34, 31 must be the furthest right one. There is then one path from 25 to 29, and we have solved!