How do I find all prime solutions $p, q, r$ of the equation $\displaystyle p(p+1)+q(q+1) = r(r+1)$? 
Find primes  $p, q, r$ of the equation $$p(p+1)+q(q+1)
= r(r+1)$$

I know that it has only one solution namely $p = q = 2,r = 3$. But i can't show that.
Thank you for any help 

May this lead to a simple proof for your problem according to your unic example of solution . There is only one
solution, namely $p = q = 2,r = 3$. To see that, we shall find all solutions
of the equation $p(p+1)+q(q+1) = n(n+1)$ where $p$ and $q$ are primes and
$n$ is a positive integer. Our equation yields
$p(p+1) = n(n+1)-q(q+1) = (n-q)(n+q+1)$,
and we must have $n > q$. Since $p$ is a prime, we have either $p|n-q$ or
$p|n+q+1$. If $p|n-q$, then we have $p\leq n-q$, which implies $p(p+1)
\leq (n-q)(n-q+1)$, and therefore $n+q+1 \leq n-q+1$, which is impossible.
Thus we have $p|n+q+ 1$, which means that for some positive integer $k$
,$n+q+1 = kp$, which implies $p+1 = k(n-q)\tag1$.
If we had $k = 1$, then $n+q+ 1 = p$ and $p+ 1 = n-q$, which gives $p-q
= n+ 1$ and $p+q = n- 1$, which is impossible, because $(p+q)-(p-q)=2q>0$ and $(n-1)-(n+1)=-2<0$. Thus, $k > 1$. From $(1)$ we
easily obtain:
$$\begin{align}
2q &= (n+q)-(n-q) \\
&= kp-1-(n-q) \\
&= k[k(n-q)-1]-1-(n-q) \\
&= (k+1)[(k-1)(n-q)-1].
\end{align}$$
Since $k \geq 2$, we have $k+1 \geq 3$. The last equality, whose left-hand side has
positive integer divisors $1, 2, q$, and $2q$ only, implies that either $k+ 1 = q$
or $k+1 = 2q$. If $k+1 = q$, then $(k-1)(n-q) = 3$, hence $(q-2)(n-q) = 3$.
This leads to either $q-2 = 1$, $n-q = 3$, that is $q = 3, n = 6, k = q-1 = 2$,
and, in view of $(1)$, $p = 5$, or else, $q-2 = 3$, $n-q = 1$, which gives $q = 5,
n = 6, k = 4$, and in view of $(1)$, $p = 3$.
On the other hand, if $k+1 = 2q$, then $(k-1)(n-q) = 2$, hence
$2(q-1)(n-q) = 2$. This leads to $q-1 = 1$ and $n-q = 1$, or $q = 2, n = 3$,
and, in view of $(1)$, $p = 2$. Thus, for positive integer $ n$, we have the
following solutions in primes $p$ and $q$: 


*

*$(p = q = 2, n = 3; 2)$,

*$ (p = 5,
q = 3, n = 6)$, and

*$(p = 3, q = 5, n = 6)$.
Only in the first solution all
three numbers are primes.
Note: If we denote by $\displaystyle t_n = \frac{n(n+1)}{2}$ the nth triangular number,
then   the equation $t_p+t_q = t_r$
has only one solution in prime numbers, namely $p = q = 2, r = 3$.
