We will find infinitely many triplets $(a,b,c)$ with $(a+b+c)^2=7(ab+bc+ca)$ and $a$, $b$, and $c$ are all odd and relatively prime. We will use the method of reverse Vieta jumping. Notice that $(a,b,c)=(3,5,41)$ satisfies. Assume we have that $(a,b,c)$ satisfies and $a<b<c$. Let $a_1< a_2$ be the two roots of $$a^2-(5b+5c)a+(b^2+c^2-5bc)$$$$a_1,a_2=\frac{5b+5c\pm \sqrt{21b^2+21c^2+70bc}}{2}$$Then $a_1=a$ and $b,c<a_2$ so we found another triplet with a higher minimum. All that is left to show is that $a_2$, $b$, and $c$ are all relatively prime and odd. As $a_1$, $b$, and $c$ were odd then $a_2$ is also odd. If then for the sake of contradiction $p|b,a_2$ for some odd prime $p$, then $$ 5c\equiv - \sqrt{21b^2+21c^2+70bc} \Rightarrow 4c^2\equiv 0 \pmod{p}\Rightarrow p|c$$which is absurd.

Take p and q are primes such that 13x7|p^2+q^2-pq and 3 does'nt divise it Let: a=19p^2+q^2-19pq b=p^2+19q^2-19pq c=p^2+q^2+17pq

lets asume $a$ is an odd number greater than $n$ if we make the values of $b,c$ equal to : $$a^2 - 2a + 2, a^2 + b^2 + ab - a - b$$the set of prime divisors of the numbers $a+b+c$ and $ab+bc+ac$ will conside

Ryan-asadi wrote: lets asume $a$ is an odd number greater than $n$ if we make the values of $b,c$ equal to : $$a^2 - 2a + 2, a^2 + b^2 + ab - a - b$$the set of prime divisors of the numbers $a+b+c$ and $ab+bc+ac$ will conside Really! In this case \[ a+b+c=(a^2 - 2 a + 4) (a^2 - a + 1), \quad ab+bc+ca=(a^2 - 2 a + 4) (a^2 - a + 1)^2. \]I think here is simple motivation: we can try to find $a, b, c$ such a for some $u, v$ we get $a+b+c=uv, ab+bc+ca=uv^2$.

How can you find those solutions?