solve for product of primes with difference of squares of primes

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I had the following two algebras:

$$C(p,q) = \frac{p^2-q^2}{4} \quad \text{and} \quad N(p,q) = pq$$

where p and q are primes greater than 10, moreover, the number of integers(i.e. the length of each prime) are equal.

So for example, if $p = 101$, then $q$ has to be a prime with exactly three digits, like $139$.

Now, I would like to find a relation between $C$ and $N$.

What I did is, I proved that $N(p,q) = pq$ is a one to one function due to the prime property and both $p$ and $q$ are greater than $10$.

Thus, if $N_{1} = N_{2}$, then $p_{1} = p_{2}, q_{1} = q_{2}$, which implies $C(p_{1}, q_{1}) = C(p_{2},q_{2})$.

Therefore, can I claim that there exist a one to one function that map $C$ to $N$?

If so, can I find a valid transformation or some formula to represent N with c?

If additional details are required, please let me know.

Thanks in advance for any efforts!