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The_Doctor
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Homework Statement
A tri-omino is an equilateral triangular tile with lines dividing the triangular face into four equilateral triangles decorated with blanks or pips. There are (N+3)(N+2)(N+1)/6 tri-ominos in a Triple N set. Is it possible to tile a triangular region without gaps or overlaps using a complete Triple N set of tri-ominos if N > 1?
Homework Equations
Pell's equation
The Attempt at a Solution
Given any number of triominoes and an equilateral triangular tile of any size, how many triominoes are needed to tile that tile? Well all equilateral triangles are similar, so let us say each triomino has a side length of n. This means the tile will have a side length of kn, due to similarity where k is arbitrary. Because we can only use whole triominoes, not fractions of them, k must be an integer. The area of each triomino is (√3/4)*n2 and the area of the tile will be (√3/4)*(kn)2. To work out how many triominoes are needed to tile the tile, we divide the tile's area by the area of each triomino. Once divided we get k^2. Thus k^2 triominoes are needed to tile an equilateral triangular tile of length kn. So we need to find out when (N+3)(N+2)(N+1)/6 is a perfect square.
(N+1)(N+2)(N+3)/6 must be in the form [itex]\frac{x^22y^23z^2}{6}[/itex]. Here is the proof: N+1, N+2 and N+3 are consecutive numbers, and thus one and only one will be divisible by 3. Secondly, because they are consecutive, N+1, N+2 and N+3 will not share any prime factors except for 2, when N+1 and N+3 is even. Because there is a 6 as the denominator, which is 2*3, (N+1)(N+2)(N+3) must have an odd power of 2 when represented in prime factors. Since only a maximum of 2 of the 3 consecutive numbers will be even and thus have 2 as its prime factors, only 1 of the three consecutive numbers can have an odd power of 2 in its prime factors. And because none of three consecutive numbers share prime factors other than 2, we can say that (N+1)(N+2)(N+3)/6 must be in the form [itex]\frac{x^22y^23z^2}{6}[/itex] and can only be in that form in order for (N+1)(N+2)(N+3)/6 to be a perfect square.
Now we don't know which of the three consecutive numbers, x^2, 2y^2 and 3z^2 will equal. So [itex]x^2-3z^2= \pm 1,\pm 2[/itex]. However,
[itex]0^2\equiv 0\pmod{3} \\
1^2\equiv 1\pmod{3} \\
2^2\equiv 1\pmod{3}[/itex]
Therefore, [itex]x^2-3z^2=1,-2[/itex]
Let's say that x^2=N+2 and 3z^2=N+1. Then x^2-3z^2=1. This gives us a Diophantine equation; specifically Pell's equation. The minimal solution to this is x=7, z=4. The solutions to this equation must also fit this equation: x^2-2y^2=-1, a negative Pell equation. Long story short, x=7 works and gives y=5. So we have 3*4^2*7^2*2*5^2/6=(4*7*5)^2
I quickly typed up a program to brute-force answers, and it only came up with N=1,47. Are these the only solutions? And how would you go about proving it?
But my method of finding the solutions is pretty bad. Is there a more logical way of finding out if there are possibilities? E.g. I just guessed if x^2=N+2, and it gave me the right answers, but is there a way of deducing it logically?