<s>[INST] Find all prime numbers satisfying... Find all prime numbers p, q, r such that 
 $p(p+1)+q(q+1)=r(r+1)$.
I tried out with $6k\pm1$ form of prime numbers but got no useful result, other methods too proved futile.
 [/INST] There is only one solution $p=q=2$ & $r=3$ , we will prove it for a positive integer $ n$
We deduce that , $p(p+1)=n(n+1)-q(q+1)=(n-q)(n+q+1)$ and we must have $ n>q$
Since $p $ is a prime we must have $ p|(n-q)$ or $p|(n+q+1)$ , Now if $p|(n-q)$ then $p\le (n-q)$ which implies $p(p+1)\le (n-q)(n-q+1)\implies (n-q)(n+q+1)\le(n-q)(n-q+1)$ and therefore $(n+q+1)\le(n-q+1)$ which is impossible and thus $p|(n+q+1)$ .
For some positive integer $k$ we have $n+q+1=kp$ & thus putting it into original equation we yeild $ \require{cancel} \cancel{p}(p+1)=k\cancel{p}(n-q)$ which is $ p+1=k(n-q)$ , For $k=1$ we have $p+q+1=n$ & $n+q+1=p$ which yields $p-n=n-p $ on subtracting which is absurd and thus $k>1$ .
Note that $kp-1=n+q,p=k(n-q)-1$ & so we have the identity $2q=(n+q)-(n-q)=kp-1-(n-q)=k(k(n-q)-1)-(n-q)=(k+1)[(k-1)(n-q)-1]$ , With these and the condition that $k\ge2\implies k+1\ge 3$ .
$2q$ has divisors $1,2,q,2q$ only which implies $k+1=q$ or $k+1=2q$ only.
If $k+1=q$ , $(k-1)(n-q)=3$ hence $(q-2)(n-q)=3$ which akes either $q-2=1$ & $n-q=3$ which yields the solutions $(p,q,k)=(5,3,2)$ and thus $n=6$ .
Else if $q-2=3$ & $ n-q=1$ which makes $p=3,q=5,n=6$ .
Now if $k+1=2q$ then $(k-1)(n-k)=2$ and $2(q-1)(n-q)=2$ which leads to $q-1=1$ & $n-q=1$ and thus $p=2,q=2,n=3$
Thus we have the following solutions in positive integer $n$ & primes $p,q$ which are $(p,q,n)=(5,3,6),(3,5,6),(2,2,3)$ and it is clear that only one prime solution exist $p=q=2,r=3$ .
Hope it helps !
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