$$(a+b)^2-c^2=ab+bc+ac\implies a+b+c=ab+bc+ac,a+b-1=c$$$2b+2a-1=ab+b(a+b-1)+a(a+b-1)\implies (a+b)^2-3(a+b)=-(ab+1)$ Obviously $-(ab+1)$ is a negative value and we know that $(a+b)^2-3(a+b)\ge \frac{-9}{2}$ So we have that $ab=3,2,1$ By testing the solutions we find that $\boxed{(a,b,c)=(1,1,1)}$ is the only solution. another approach Quote: a+b+c=ab+bc+ac $$\sum_{cyc}a(b-1)=0$$clearly $\boxed{(a,b,c)=(1,1,1)}$ now if we have $(a,b,c)>1$ let $a=x+1,b=y+1,c=z+1$ now we have $(x+1)y+(y+1)z+(z+1)x=0 \Rightarrow\Leftarrow $ since $\min\{(x+1)y+(y+1)z+(z+1)x\}=6$ so the solution is $\boxed{(a,b,c)=(1,1,1)}$

$(a+b)^2=(a+c)(b+c)=ab+bc+ca+c^2$. $ab+bc+ca=(a+b)^2-c^2=(a+b-c)(a+b+c).$ Since $a+b-c<a+b+c$ (by positivity), and $ab+bc+ca$ is prime, it follows that $a+b-c=1 \implies a+b=c+1$. So, $ab+bc+ca=a+b+c=2c+1 \implies ab-1=c(2-(a+b))=c(1-c).$ If $c>1$, the L.H.S $=c(1-c)<0$, but $ab-1 \geq 0 \forall a, b \in \mathbb{Z}^+$ (contradiction). Thus, $\boxed{c=1}$ and $a+b=1+1=2 \implies \boxed{a=1}, \boxed{b=1}$. Therefore, $\boxed{(a,b,c)=(1,1,1)}$ is the only solution.

We start with $\frac{a+b}{a+c}=\frac{b+c}{b+a}\Longleftrightarrow (a+b)^2=(b+c)(a+c)$ Claim: $(a+b)^2=(a+c)(b+c)\Longleftrightarrow a=b=c$ Proof: We use the following substitution, $x=a-c$ and $y=b-c$ Then $(a+b)^2=(a+c)(b+c)~\Longleftrightarrow (x+y+2c)^2=(x+2c)(y+2c)$ $\qquad\qquad\qquad\qquad\qquad\qquad\Longleftrightarrow (x+y+2c)^2-(x+2c)(y+2c)$ $\qquad\qquad\qquad\qquad\qquad\qquad\Longleftrightarrow x^2+y^2+4c^2-2(xy+2yc+2cx)-xy-2xc-2cy-4c^2=0$ $\qquad\qquad\qquad\qquad\qquad\qquad\Longleftrightarrow x^2+y^2+xy=0$ and since $x,y\ge 0$ then $(x,y)=(0,0)$ is the only solution which gives : $(a-c=0\Leftrightarrow a=c$ and $b-c=0 \Leftrightarrow b=c)\Longleftrightarrow a=b=c \quad \square$ Finding $a,b,c$ : Let's find all $(a,b,c)$ such that $ab+bc+ca\in\mathbb{P}$. Let $n\in\mathbb N$ s.t. $a=b=c=n$ then $ab+bc+ca=3n^2$ and since $3\in\mathbb P$ then $n=1$ is the only solution in $\mathbb R_{\ge 0}$ Therefore, $(a,b,c)=(1,1,1)$ is the only solution.

Compute $gcd$ of $(a,b) , (b,c) , (a,c)$ and of cyclic sums the result follows.

alba_tross1867 wrote: Compute $\gcd$ of $(a,b) , (b,c) , (a,c)$ and of cyclic sums the result follows. prestigious