# A semiprime proof

• Nov 15th 2009, 10:57 PM
Bacterius
A semiprime puzzle
Hi all, here is a little conjecture (I do have my proof but it could be interesting to see others' proofs), let's see how you do.

Say $\displaystyle n = pq$ with $\displaystyle p$ and $\displaystyle q$ primes ($\displaystyle n$ is a semiprime, thus).

Prove that there exists only two distinct solution pairs $\displaystyle (a, m)$ such that $\displaystyle m^2 - 4n = a^2$ with $\displaystyle m \in N$, $\displaystyle a \in N$.

Good luck !
• Nov 18th 2009, 01:04 AM
simplependulum
Are the solutions of $\displaystyle (a,m)$ $\displaystyle ( p-q , p+q ) , ( q-p ,p+q)$ ?
• Nov 18th 2009, 01:22 AM
Bacterius
Nearly : the solutions $\displaystyle (a;m)$ are $\displaystyle (n - 1;n + 1)$ and $\displaystyle (q - p; p + q)$
• Nov 20th 2009, 02:29 PM
pouncep
i think simplependulum is right
i got the same answer as simplependulum

• Nov 20th 2009, 03:17 PM
Bacterius
What to you mean by $\displaystyle 2pq$ ?
Anyway the point is to prove it, not give the two answers (Sweating)
• Nov 28th 2009, 11:32 PM
roninpro
Hello. I noticed that you didn't actually receive a (formal) solution, so I thought that I would give mine.

I will rearrange the equation in question:

$\displaystyle m^2-a^2=4n$

And once more:

$\displaystyle (m + a)(m - a) = 4pq$

We know for sure that the terms $\displaystyle (m+a)$ and $\displaystyle (m-a)$ are divisors of $\displaystyle 4pq$. Denoting $\displaystyle d_1,d_2$ as arbitrary divisors of $\displaystyle 4pq$, it suffices to solve all of the systems of the form $\displaystyle m+a=d_1$; $\displaystyle m-a=d_2$.

The systems have integer solutions precisely when $\displaystyle d_1=2pq,d_2=2$ and $\displaystyle d_1=2p,d_2=2q$. In particular, $\displaystyle m=pq+1,a=pq-1$ and $\displaystyle m=p+q,a=p-q$, respectively. (Note that it may be necessary to reverse the roles of $\displaystyle d_1,d_2$ to adjust the sign of $\displaystyle a$.)

Other combinations do not work for any semiprime or lead to redundancy when one of the factors is 2. For example, $\displaystyle d_1=4,d_2=pq$ gives $\displaystyle m=\frac{4+pq}{2},a=\frac{4-pq}{2}$. We can see that $\displaystyle a$ is always nonpositive, etc...

I think that this does it, for the most part.

Thanks for the problem.
• Nov 30th 2009, 02:51 AM
Bacterius
Interesting, indeed. My proof uses the fact that this equation can be considered as being the discriminant of the quadratic equation $\displaystyle p^2 - mp + n = 0$, where $\displaystyle m = p + q$. Therefore I can easily prove that there are only two solutions which are ... etc ...
Good idea though, wouldn't have seen it like that (Rock)