F (x) defined on a set of positive integers is a positive integer for any m, N, where f (M + n) = f (m) + F (n) + 4 (M + n) - 2, and f (1) = 1 [1]

F (x) defined on a set of positive integers is a positive integer for any m, N, where f (M + n) = f (m) + F (n) + 4 (M + n) - 2, and f (1) = 1 [1]

Let m = x, n = 1, we get f (x + 1) = f (x) + 4x + 3
f(2)=f(1)+4*1+3
f(3)=f(2)+4*2+3
f(4)=f(3)+4*3+3
.
f(x)=f(x-1)+4*(x-1)+3
Add up,
f(x）=f(1)+4*（1+2+3+...+(x-1))+3*(x-1)=2x²+x-2
Obviously, the minimum value of F (x) is 1,
So M & sup2; - TM-1 ≤ 1 holds for any m ∈ [- 1,1]
When m = 0, the inequality of t ∈ r holds;
When m ＜ 0, the original formula is equivalent to that t ≤ m-2 / M is constant when m ∈ [- 1,0], and the minimum value of function m-2 / M is 1 (the function is single increasing function), so t ≤ 1;
When m > 0, the original formula is equivalent to that t ≥ m-2 / M holds constant in M ∈ (0,1), and the maximum value of m-2 / M is - 1 (the function is a simple increasing function), so t ≥ - 1
In conclusion, when - 1 ≤ m < 0, t ≤ 1
When m = 0, t ∈ R
When 0 ＜ m ≤ 1, t ≥ - 1

Define an operation "*": 1 * 1 = 2, m * n = k, m * (n + 1) = K + 3 (m, N, K are positive integers), then the result of 1 * 2007 is

1*1=2
1*2=2+3=5
1*3=5+3=8
...
1*2007=2007*3-1=6020

If M = 2006 ^ 2 + 2006 ^ 2 * 2007 ^ 2 + 2007 ^ 2
It is proved that M is a complete square number and an odd number

M is obviously odd
m=(2007-1)^2+[(2007-1)2007]^2+2007^2
=2007^2-2*2007+1+(2007^2-2007)^2+2007^2
=2007^4-2*2007^3+3*2007^2-2*2007+1
=2007^4+2*2007^2+1-2*2007(2007^2+1)+2007^2
=(2007^2+1)^2-2*2007(2007^2+1)+2007^2
=(2007^2+1-2007)^2