1. ## 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 !

2. Are the solutions of $(a,m)$ $( p-q , p+q ) , ( q-p ,p+q)$ ?

3. Nearly : the solutions $(a;m)$ are $(n - 1;n + 1)$ and $(q - p; p + q)$

4. ## i think simplependulum is right

i got the same answer as simplependulum

5. What to you mean by $2pq$ ?
Anyway the point is to prove it, not give the two answers

6. 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.

7. 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