I think I got it, although I don't know if there is a simpler way to do this:
f(xy) gcd(f(x), y) = f(x) f(y) and f(yx) gcd(f(y), x) = f(y) f(x) imply gcd(f(x),y) = gcd(f(y),x) for all x,y.
Two important properties:
[1] If we let f(x)=n, then n=gcd(f(x),n)=gcd(f(n),x) implying n divides x (notation: n|x). That is, f(x)|x for all x.
[2] Note that gcd(f(x),y)|y and gcd(f(y),x)|x. If gcd(x,y)=1, it follows that gcd(f(x),y)=gcd(f(y),x)=1. Then f(xy) gcd(f(x), y) = f(x) f(y) becomes f(xy) = f(x) f(y). Therefore f(x) must be a
multiplicative function.
Now we check specific values. First, by property [1], f(1) must be 1.
Then we check f(p) for prime p. By property [1], f(p) is either 1 or p:
Case 1: f(p)=1. Then x=y=p gives f(p
2) gcd(f(p),p) = f(p)f(p) ---> f(p
2)=1. Likewise, x=p
2, y=p gives f(p
3)=1, and so on. By induction, f(p
n)=1.
Case 2: f(p)=p. Then x=y=p gives f(p
2) gcd(f(p),p) = f(p)f(p) ---> f(p
2)=p. Likewise, x=p
2, y=p gives f(p
3)=p, and so on. By induction, f(p
n)=p.
Thus, for each prime p, we have a choice whether to assign f(p
n)=1 for all n≥1, or f(p
n)=p for all n≥1; that is, whether to include p as a prime factor in the output or not. Together with with the multiplicative function property [2], we can summarize possible f as follows: Any function satisfying f(xy) gcd(f(x), y) = f(x) f(y) must be of the form f
Q(x), where Q is a (possibly infinite) subset of the primes, and f
Q(x) is the product of all prime factors of x that are in Q.
----
There is one other thing: The above reasoning does not actually verify that f=f
Q(x) satisfies the equation. To verify it, we need one more thing:
Consider a function f=f
Q(x). To check that f
Q(xy) gcd(f
Q(x), y) = f
Q(x) f
Q(y), we will look at each prime p:
Case 1: p is not in Q: None of the factors in the equation will contain the factor p then.
Case 2: p is in Q but divides neither x nor y: None of the factors in the equation will contain the factor p then.
Case 3: p is in Q and divides x, but not y: The factors f
Q(xy) and f
Q(x) will each contain exactly one factor of p, and only these factors.
Case 4: p is in Q and divides y, but not x: The factors f
Q(xy) and f
Q(y) will each contain exactly one factor of p, and only these factors.
Case 5: p is in Q and divides both x and y: The factors f
Q(xy), gcd(f
Q(x), y), f
Q(x) and f
Q(y) will each contain exactly one factor of p.
In all cases, the factors of p are balanced on both sides. Doing this over all primes p shows that the equation holds.
There might be a problem in considering the yellow and gray regions adjacent, considering the purple and pink regions at the same "corner" are equally as close! (The regions to which the four-color theorem apply shouldn't be considered touching at "corners".)
We form a weaving loom pattern in the unit square as follows:
* On the left side from the top-left corner to the bottom-left corner are n equally-spaced points A1, A2, ..., An.
* On the bottom side from the bottom-left corner to the bottom-right corner are n equally-spaced points B1, B2, ..., Bn. Note that B1 = An.
* Draw n line segments: from A1 to B1, A2 to B2, and so on, up to An to Bn.
The figure above shows n=100.
Question: As n goes to infinity, what is the curve formed by the upper edge of the pattern?
----
Fiddler on the proof also has a bonus question ("Extra Credit"). Rotate the above pattern around so it looks like this:
Question: As n goes to infinity, what fraction of the unit square is not covered by the four patterns (that is, what is the area of the central white region)?
Looks like 入ルナ to me (meaning DON'T ENTER) and it's at the point near the very end where you are blocked by a wall you have to glitch through (almost impossible even as big Mario).