Let n=0. Then F(0) = c. Which implies c = 0 (mod m).
Let n=-1. Then F(-1) = a-b+c-1. Which implies a = b+1 (mod m).
Let n=1. Then F(1) = a+b+c+1. Which implies a = -b-1 (mod m).

Combining these results implies 2a = 0 (mod m) and 2b = -2 (mod m).

Now let n=2. Then F(2) = 8 + 4a + 2b + c. Combined with the above results yields 6 = 0 (mod m). m must be a factor of 6.

The largest factor of 6 is itself. One specific polynomial F(x) = x^3 + 6x^2 + 11x + 6 = (x+1)*(x+2)*(x+3) satisfies the requirements stated in the problem. Therefore the largest m is 6.

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