# A semiprime proof

Printable View

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

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

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

your first answer will get 2pq
• Nov 20th 2009, 03:17 PM
Bacterius
What to you mean by $2pq$ ?
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:

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

And once more:

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

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

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

Other combinations do not work for any semiprime or lead to redundancy when one of the factors is 2. For example, $d_1=4,d_2=pq$ gives $m=\frac{4+pq}{2},a=\frac{4-pq}{2}$. We can see that $a$ 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 $p^2 - mp + n = 0$, where $m = p + q$. 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)