from apmc

[the arrivals]

find triples (p,q,n) with p,q primes and n an integer which is greater or equal 1
satisfying the awkward( ) relation
p(p+1)+q(q+1)=n(n+1)
the arrivals

[Moon]

Nice problem:
Hint (I'll post complete solution later)

Use some inequalities to impose some bounds. You should get $(p,q)=(5,3)$.

[Avik Roy]

Some basic number theory also performs well

Try with $q(q+1) = (n-p)(n+p+1)$
and don't forget: q is a PRIME

[the arrivals]

(2,2,3) perhaps a solution also.

[Moon]

Yup...the solutions are $(p,q,n)=(2,2,3),(5,3,6)$.

Case 1: $p=q$
We have $2p(p+1)=n(n+1)$.

Case 1(a): $p | n$
So, $n \geq 2p \iff n+2 \geq 2(p+1)$ ($n=p$ is not possible). We also have $2(p+1)=k(n+1)$. So, $k(n+1) \leq n+2$. Checking some values of $k$ we can see that this case is not possible.

Case 1(b): $p |(n+1)$
So, $(n+1) \geq 2p \iff n+3 \geq 2(p+1)=kn \iff (k-1)n \leq 3 $. Only $k=1$ is possible. We have $(p,q,n)=(2,2,3)$.

Case 2: $p >q$
The equation rearranges to $p(p+1)=(n-q)(n+q+1)$.

Case 2(a): $p | (n-q)$
Then $ p \leq (n-q) \iff p+1 \leq n-q+1 < n+q+1$. Contradiction!

Case 2(b): $p | (n+q+1)$
As $n-q< n+q+1$, $n+q+1 \geq 2p$. Then we have $n-q <p \iff n+q+1 <p+2q+1 <3p$.

Finally we have $3p>n+q+1 \geq 2p \iff n+q+1=2p$. So, $2(n-q)=p+1$. From these two equalities we have $3n-5q=3$. So $3 |q$ and recall that $q$ is a prime. So $q=3$. From the equalities we also have $4q+3=3p$; so, $p=5$.
We are done!

Remark: When writing more official solution you can not just write, "it is easy to check that..." you must complete write how to check as well.

[Avik Roy]

I just got a combinatorial proof of the problem
The given relation can be written as: \[\begin{pmatrix}
n+1\\2

\end{pmatrix} = \begin{pmatrix}
p+1\\2

\end{pmatrix} +\begin{pmatrix}
q+1\\2

\end{pmatrix}\]
Now, since $p<n$, lets consider that we choose $n-p$ elements from available $n+1$ elements and keep them aside in choosing elements. So we have $p+1$ elements to choose from. In the following cases, we choose elements from the kept aside $n-p$ elements in all possible combinations to keep or not to keep them in our selection to complete the quota. However, we have only $2$ elements to select, so $n-p$ can't exceed $2$. Choosing $n-p=2$, we have the following equation to satisfy:$\begin{pmatrix}
p+1\\2

\end{pmatrix}+2\begin{pmatrix}
p+1\\1

\end{pmatrix}+1=\begin{pmatrix}
n+1\\2
\end{pmatrix}$
and putting $n=p+2$ here leads one to complete gibberish. Hence $n=p+1$ is to satisfy. Comaring the given equation with the identity: \[\begin{pmatrix}
n+1\\r

\end{pmatrix} = \begin{pmatrix}
n\\r

\end{pmatrix} +\begin{pmatrix}
n\\r-1

\end{pmatrix}\]
we hereby obtain
$\begin{pmatrix} q+1\\2 \end{pmatrix} = \begin{pmatrix} p+1\\1 \end{pmatrix}$
$\Rightarrow q(q+1) = 2(p+1)$
Now, we have to consider the following cases-
C1: $q=2$ which leads to the solution $(2,2,3)$
C2: $q|p+1$ implying the following set of equations:
$p+1 = kq$
$q+1 = k.2$
Adding them we receive-
$p+q+2 = k(q+2)$
$\Rightarrow k = 1 + \frac {p}{q+2}$
since $p$ is a prime $p=q+2$ and $k=2$
This leads us to the solution $(5,3,6)$

[the arrivals]

oow!! awesome!! what make you approching like this?? very cool instead.appreciating.

[Avik Roy]

@the arrivals, whom did you address??

[Moon]

The approach is really unorthodox and cool!
However, to me these explanations are not clear.

Now, since $p<n$, lets consider that we choose $n-p$ elements from available $n+1$ elements and keep them aside in choosing elements. So we have $p+1$ elements to choose from. In the following cases, we choose elements from the kept aside $n-p$ elements in all possible combinations to keep or not to keep them in our selection to complete the quota. However, we have only $2$ elements to select, so $n-p$ can't exceed $2$. Choosing $n-p=2$

BTW you need not use \pmatrix, you may use \binom or \dbinom
I mean $\dbinom{n+1}{2}$

[the arrivals]

nobody just myself @avik roy