Let Q' denote the set of all positive rational numbers. If f:Q'→Q' satisfies
f(x^{2}f(y)^{2})=f(x)^{2}f(y),
evaluate f(2019).
The square root function would also work with this equation. But whoops, it is not rational! Can that get us anywhere? Let's see...
Call the original equation "I":
f( x^2 f(y)^2 ) = f(x)^2 f(y)
With x=1, this becomes
II: f( f(y)^2 ) = c^2 f(y)
where c:=f(1). This almost looks like the square root function... Likewise, with y=1 we get
III: f(c^2 x^2) = c f(x)^2
Let's figure out the constant c. When we set x=1/c in III, we get
c = f(1) = f(c^2 (1/c)^2) =^{III} c f(1/c)^2 => f(1/c) = 1
and setting y=1/c in II gives
f( f(1/c)^2 ) = c^2 f(1/c) = c^2
The left side is f(1^2) = c and thus c=1.
So II and III become a little handier (also replace y by x in II):
II: f( f(x)^2 ) = f(x)
III: f( x^2 ) = f(x)^2
Now plug this into I again:
f( x^2 f(y)^2 ) = f( (x f(y))^2 ) = ^{III} f( x f(y) )^2 = ^{I} f(x)^2 f(y)
With x = f(y) the last equation becomes
f( f(y)^2 )^2 = f( f(y) )^2 f(y)
Using II on the left side gives
f(y)^2 = f( f(y) )^2 f(y) <=> f(y) = f( f(y) )^2
So for any w in the image of f, i.e. w=f(y) for some y, we have
w = f(w)^2
or
IV: f(w) = sqrt(w)
What does this mean? Well, because f is rational, we can repeatedly apply f to w (i.e. take square roots) and we keep getting rational numbers. The only w you can do this with is w=1. Hence f(y) = 1 for all rational y, so f(2019) = 1.
I am afraid the solution has not quite level3 length, but this is the shortest I could come up with.
Edited on November 27, 2019, 3:58 pm

Posted by JLo
on 20191127 15:51:00 