no, it doesn't exist. we have: A B C D E F G H I. $ f_n$ is the greatest number that we use. CASE 1)$ f_n$ is in A (or simmetrically in C,G,I) we must have A+B+C=C+F+I so A+B=C+I but C+I $ \le$ A < A+B, absurd. CASE 2)$ f_n$ is in B(or simmetrically in D,F,H) we must have B+E+H=D+E+F so B+H=D+F but D+F $ \le$ B < B+H, absurd. LAST CASE)$ f_n$ is in E. we must have G+H+I=A+B+C=D+E+F>E. but $ \frac{(G+H+I)+(A+B+C)}{2} \le \frac{\sum_{i=n-6}^{n-1}{f_i}}{2}=\frac{(f_{n-1}+f_{n-2})+(\sum_{n-6}^{n-3}{f_i})}{2}< f_n$, absurde bye paolo

an easier approach in which we prove that even the sum of 3 different rows cannot be the same; Let f(n+4) be the largest occuring Fibonacci number. Then there must be a row in which the largest Fibonacci number is no larger than f(n+2). But the sum of that row is no larger than f(n) + f(n+1) + f(n+2) = f(n) + f(n+3) < f(n+2) + f(n+3) = f(n+4), impossible.

Note that the previous proof alsof easily generalizes to m*m magic squares; If we let f(m+n-1) be the largest occuring Fibonacci number, there must be a row in which the largest Fibonacci number is no larger than f(n). The sum in that row is smaller than or equal to: f(n-m+1) + f(n-m+2) + .. + f(n) < (m-1)f(n) < 1.5^(m-1)f(n) < f(n+m-1), contradiction. Another approach; Let X = F(x) = the x-th fibonacci number. Let A B C D E F G H I be our 3*3 magic square. Then A + I = 2E = C + G. This implies that max(A, C, G, I) is at least f(e+2). But this is already larger than 2E, contradiction.