On f(x^2 + y^2) = f(x)^2 + f(y)^2

Harold Chicot asked for the most general integer function f(N) such
that f(1) is positive and

            f(m^2 + n^2)  =  f(m)^2 +  f(n)^2                  (1)

Francois Dress pointed out that the only such function is the identity
function f(n) = n.  First, he notes by inspection that f(k) = k for 
all k up to 5.  Then, notice that any integer n greater than 5 can be 
written uniquely in the form n = 5a-2b with b=0,1,2,3, or 4.  (For 
example, 6 is given by setting a=2 and b=2.)  Now we note the identity

           (5n-2a)^2 + (a)^2  =  (3n-2a)^2 + (4n-a)^2  

Applying the function f to both sides and using the property (1), we
have
          f(5n-2a)^2 + f(a)^2  =  f(3n-2a)^2 + f(4n-a)^2  

Since each of the quantities 3n-2a, 4n-a, and a is less than 5n-2a
(with a=0,1,2,3, or 4), and since we have established that f(k)=k for 
all k less than 5n-2a, we can expand three of the terms in the above
equality to give

    f(5n-2a)^2  =  (3n-2a)^2 + (4n-a)^2 - a^2

                =   9n^2 - 12na + 4a^2  + 16n^2 - 8na + a^2 - a^2

                =   25n^2 - 20na + 4a^2

                =   (5n-2a)^2

Therefore, f(5n-2a) = 5n-2a, and by induction this proves that f(n)=n
for all n.
 
This nice proof makes me wonder if it can be made to work if we allow 
f(k) to take on REAL values.  Let's see...the value of f(0) is forced 
by the relation 
                  f(0) = f(0)^2 + f(0)^2 

to be either 0 or 1/2.  If we let f(0)=1/2 then we have a quadratic
in f(1)
                  f(1) = 1/4 + f(1)^2

which implies that f(1) also equals 1/2.  Then we have the equation

                  f(2) = f(1)^2 + f(1)^2 = 1/2

From here it's clear that setting f(k)=1/2 for all k gives a consistent
solution, and this is the ONLY other solution besides f(k)=k (which, as 
FD showed, is the only integer solution).  I'll leave it as an exercise 
for the reader to show that this implies all the complex roots of the 
zeta function must have real part 1/2.