Is There a Function f: Z -> Z Such That f(f(n))=-n for Every Integer?

[ypatia]

Is there any function (if any) f: Z -> Z such that
f(f(n))=-n , for every n belongs to Z(integers) ??


I think that there is not any function like the one described above but how can we prove it. Any ideas??
Thanks in Advance

 

[mathman]

How about f(n)=in?

 

[CRGreathouse]

Not integer-valued. (I assume if the OP meant Gaussian integers that would have been mentioned, since that's the obvious solution.)

I've been thinking about this for a few hours now and I can't see any way to do it, but I can't prove that it's impossible.

 

[chronon]

How about

for n>0
f(2n-1)=2n
f(2n)=-2n+1
f(-2n+1)=-2n
f(-2n)=2n-1

f(0)=0

 

[AUMathTutor]

Nice, chronon. Nice.

 

[matt grime]

Indeed - it's nice to visualize f as a piecewise permutation

(0)(-2,-1,2,1)(-4,-3,4,3)...(-2n,-2n+1,2n,2n-1)...

and recall that (abcd)^2=(ac)(bd)