Let Q' denote the set of all positive rational numbers. If f:Q'→Q' satisfies
f(x2f(y)2)=f(x)2f(y),
evaluate f(2019).
f(x)=1 is a solution
f(x^2*f(y)^2)=1
and
f(x)^2*f(y)=1^2*1=1
not able to find a proof that this is the only solution
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Posted by Daniel
on 2019-11-01 10:17:10 |