# solve for product of primes with difference of squares of primes

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?

• You have not specified any ordering restriction on $p$ and $q$ relative to each other. Thus, note that if $p_1 \neq q_1$, then if $N_1 = N_2$, you can conclude that in addition to $p_1 = p_2, q_1 = q_2$ you stated, it's also possible that $p_1 = q_2, q_1 = p_2$. Mar 27 '19 at 1:56
• @JohnOmielan Hey John! Thanks for replying, the only restriction I could have on $p$ and $q$ is they do not equal to each other and I usually put $p>q$, btw, how did you get $p_{1} = q_{2}$ and $q_{1} = p_{2}$ in addtion to the original statement? Mar 27 '19 at 2:09
• @PetaGlz If $p \gt q$, then what I wrote doesn't apply. The reason it could apply without ordering is that when you factor $N_1$ to get $2$ prime factors, you can assign them in either order to $p_1$ and $q_1$. For example, $N = 15$ gives primes $3$ and $5$ as prime factors, allowing $(p,q)$ to be $(3,5)$ or $(5,3)$. Mar 27 '19 at 2:13