Find a function over natural numbers with specific properties I am searching for a function F, over the natural numbers, such that the following properties are true:
1) F is one-to-one, each input has a unique output. 
F(x) can be competed in a polynomial speed relative to the number of digits in x. (No summations with sigma, we only use simple operations such as mod and addition that are rapidly found).
2) There exists an operation over N, let it be called OP. 
OP takes two natural numbers and returns one number. It must also be computable in polynomial time relative to the number of digits of its input.
3) If a+b=c, then F(a) OP F(b)=F(c), and if a+b not equal to c, then F(a) OP F(b) not equal to F(c)
4) CRITICALLY, if a+b>c, we cannot predict whether F(a) OP F(b) is greater or less than F(c) using only the numerical magnitude of a and b and c
5) Desired but not necessary: F can be parametrised with a number y,  and we obtain several versions of F. But we cannot predict the comparison of F(a) OP F(b) to F(c) on a single parameter y using the result on other parameter y's. Examples of functions that take parameters: raising numbers to powers, where y is the power to be raised to.
I experimented with modulos and powers, but my knowledge of number theory is far inferior to the required level to produce meaningful results.

Suppose that we have an injective function $F:\mathbb{N}\rightarrow\mathbb{N}$.  Because the properties of $\oplus$ (the "OP") are checked only on $F(\mathbb{N})\times F(\mathbb{N})$, then its definition is not important outside of this set, so just define $F(a) \oplus F(b) = F(c)$: OP is already fully defined in your question (given $F$).
The core of the problem is to find $F$: $F(x)=x$ doesn't work, but for example $F(x)=2x$ if $x$ is even, $F(x)=yx$ if $x$ is odd, works when $y$ is an odd parameter, in some sense: if $y=3$, then $7<8$ but $F(7)=21 > 16 = F(8)$.
