The expression can be written as $(2p+1)^2+(2q+1)^2=(2n+1)^2+1$ or $(2p+1)^2-1=(2n+1)^2-(2q+1)^2$ or $p(p+1)=(n+q+1)q$ and $q(q+1)=(n+p+1)p$ so $p|q+1$ and $q|p+1$ so $p\le q+1\le p+2$ and $q\le p+1\le q+2$ if $p=q+1$ then p=3 and q=2 and if $q=p+1$ then q=3 and p=2 considering any of them we have $n(n+1)=3*4+2*3=18$ but no integer soln for n in both cases so $p=q$ then $n(n+1)=2p(p+1)$ so $p(p+1)=n(n+1)/2=1+2+...+n$

Finalisation of Aniv's idea: So we get: $p\left| q+1 \right.\text{ (1) and }q\left| p+1\text{ (2)} \right.$ 1) If $p=q$, we get from $(1)$ or $(2)$ that $p\left| p+1 \right.$, contradiction. 2) So $q\ge p+1$ 3) If $q>p+1$, from $(2)$, we get contradiction. 4) So $q=p+1$, thus from $(1)$ we get that $p\left| p+2\Rightarrow p=2\Rightarrow q=3 \right.$ So only $(p,q)$ solutions there are $(2,3), (3,2)$, but this not fit to the proposed equation. Done

TuZo wrote: Finalisation of Aniv's idea: So we get: $p\left| q+1 \right.\text{ (1) and }q\left| p+1\text{ (2)} \right.$ 1) If $p=q$, we get from $(1)$ or $(2)$ that $p\left| p+1 \right.$, contradiction. 2) So $q\ge p+1$ 3) If $q>p+1$, from $(2)$, we get contradiction. 4) So $q=p+1$, thus from $(1)$ we get that $p\left| p+2\Rightarrow p=2\Rightarrow q=3 \right.$ So only $(p,q)$ solutions there are $(2,3), (3,2)$. Done Thank you because I was stuck at that point!! But (2,3) or (3,2) won't give any integer value for $n$ As this will lead to $n^2+n=18$ Or $(2n+1)^2=4*18+1=73$ so no solution exists And it can be also said In other way If $p=q$ Then the relation becomes $p(p+1)=(n+p+1)(p)$ Or $n=0$ but as n is natural number so $p\ne q$ and then remaining will be same

The only solutions of the Diophantine equation $(1) \;\; (p + 1) + q(q + 1) = n(n+1)$ in primes $p,q$ and positive integers $n$ are $(p,q,n) = (2,2,3), (3,5,6), (5,3,6)$. Proof: We may WLOG assume (since $p$ and $q$ are symmetric in equation (1)) assume $p \geq q$. Equation (1) is equivalent to $(2) \;\; p(p + 1) = (n - q)(n + q + 1)$. Hence according to equation (2) and the fact that $p$ is a prime we have the following two cases to consider: Case 1: $p | n - q$. Then there is a positive integer $r$ s.t. $n =pr + q$, which inserted in equation (2) result is $(3) \;\; p + 1 = r(pr+ 2q + 1)$. The fact that in equation (3) the $LHS \geq p + 2q + 1 \geq p + 5$ implies equation (1) has no solution by contradiction. Case 2: $p | n + q + 1$. Then there is a positive integer $t$ s.t. $(4) \;\; n = pt - q - 1$, which inserted in equation (2) result is $(5) \;\; p + 1 = t(pt - 2q - 1)$. Consequently $t | p + 1$, yielding $(6) \;\; p = st - 1$, where $s$ is a positive integer, which inserted in equation (5) give us $s = t(st - 1) - 2q - 1$, or alternatively $(7) \;\; 2q = (t + 1)[s(t - 1) - 1]$. Now $p \geq q$ yields $(t^2 - 1)s - (t + 1) \leq 2st - 2$ by equations (6)-(7), i.e. $(8) \;\; (t^2 - 2t - 1)s < t - 1$. Assume $t \geq 3$. Then $x = t - 1 \leq 3 - 1 = 2$, which according to inequality (8) means ${\textstyle (9) \;\; s \leq \frac{x}{x^2 - 2} = f(x)}$. The fact that ${\textstyle f^{\prime} (x) = -\frac{x^2 + 2}{(x^2 - 2)^2} < 0}$ combined with inequality (9) yields ${\textstyle s \leq f(2) = \frac{2}{2^2 - 2} = \frac{2}{4 - 2} = \frac{2}{2} = 1}$. Hence the only possibility is $s=1$ and $x=2=t-1$, i.e. $(s,t) = (1,3)$, which inserted in equations (6) and (7) give us $p=q=2$. Thus according to equation (4) we obtain $n = pt - q - 1 = 2 \cdot 3 - 2 - 1 = 6 - 3 = 3$. Finally assume $t \leq 2$. If $t=1$, then $q=-1$ by equation (7). This contradiction implies $t=2$, which inserted in equation (7) give us $(10) \;\; 2q= 3(s - 1)$, yielding $3 | q$. Hence $q=3$ (since $q$ is a prime), yielding $s=3$ by equation (10). Therefore $p = st - 1 = 3 \cdot 2 - 1 = 6 - 1 = 5$ by equation (6), which according to equation (4) means $n = pt - q - 1 = 5 \cdot 2 - 3 - 1 = 10 - 4 = 6$. Conclusion: The only solutions of equation (1), where $p,q$ are primes and $ n$ is a positive integer, are $(p,q,n) = (2,2,3), (3,5,6), (5,3,6)$. q.e.d.

Solar Plexsus wrote: The only solutions of the Diophantine equation $(1) \;\; (p + 1) + q(q + 1) = n(n+1)$ in primes $p,q$ and positive integers $n$ are $(p,q,n) = (2,2,3), (3,5,6), (5,3,6)$. Proof: We may WLOG assume (since $p$ and $q$ are symmetric in equation (1)) assume $p \geq q$. Equation (1) is equivalent to $(2) \;\; p(p + 1) = (n - q)(n + q + 1)$. Hence according to equation (2) and the fact that $p$ is a prime we have the following two cases to consider: Case 1: $p | n - q$. Then there is a positive integer $r$ s.t. $n =pr + q$, which inserted in equation (2) result is $(3) \;\; p + 1 = r(pr+ 2q + 1)$. The fact that in equation (3) the $LHS \geq p + 2q + 1 \geq p + 5$ implies equation (1) has no solution by contradiction. Case 2: $p | n + q + 1$. Then there is a positive integer $t$ s.t. $(4) \;\; n = pt - q - 1$, which inserted in equation (2) result is $(5) \;\; p + 1 = t(pt - 2q - 1)$. Consequently $t | p + 1$, yielding $(6) \;\; p = st - 1$, where $s$ is a positive integer, which inserted in equation (5) give us $s = t(st - 1) - 2q - 1$, or alternatively $(7) \;\; 2q = (t + 1)[s(t - 1) - 1]$. Now $p \geq q$ yields $(t^2 - 1)s - (t + 1) \leq 2st - 2$ by equations (6)-(7), i.e. $(8) \;\; (t^2 - 2t - 1)s < t - 1$. Assume $t \geq 3$. Then $x = t - 1 \leq 3 - 1 = 2$, which according to inequality (8) means ${\textstyle (9) \;\; s \leq \frac{x}{x^2 - 2} = f(x)}$. The fact that ${\textstyle f^{\prime} (x) = -\frac{x^2 + 2}{(x^2 - 2)^2} < 0}$ combined with inequality (9) yields ${\textstyle s \leq f(2) = \frac{2}{2^2 - 2} = \frac{2}{4 - 2} = \frac{2}{2} = 1}$. Hence the only possibility is $s=1$ and $x=2=t-1$, i.e. $(s,t) = (1,3)$, which inserted in equations (6) and (7) give us $p=q=2$. Thus according to equation (4) we obtain $n = pt - q - 1 = 2 \cdot 3 - 2 - 1 = 6 - 3 = 3$. Finally assume $t \leq 2$. If $t=1$, then $q=-1$ by equation (7). This contradiction implies $t=2$, which inserted in equation (7) give us $(10) \;\; 2q= 3(s - 1)$, yielding $3 | q$. Hence $q=3$ (since $q$ is a prime), yielding $s=3$ by equation (10). Therefore $p = st - 1 = 3 \cdot 2 - 1 = 6 - 1 = 5$ by equation (6), which according to equation (4) means $n = pt - q - 1 = 5 \cdot 2 - 3 - 1 = 10 - 4 = 6$. Conclusion: The only solutions of equation (1), where $p,q$ are primes and $ n$ is a positive integer, are $(p,q,n) = (2,2,3), (3,5,6), (5,3,6)$. q.e.d. I am very sorry for the mistake I made And sorry to TuZo for his effort on a wrong approach

Switzerland Final Round 2016