The only triple of positive integers $(a,b,c)$ satisfying the equations $(1) \;\; GCD(a^2,b^2) + GCD(a,bc) + GCD(b,ac) + GCD(c,ab) = 239^2 = ab + c$. is $(a,b,c) = (238,238,477)$. Proof:. Let $d=GCD(a,b)$. Then there are two coprime positive integers $s$ and $t$ s.t. $(a,b) = (ds,dt)$. Consequently by equations (1) $(2) \;\; d^2 + d \cdot GCD(s,c) + d \cdot GCD(t,c) + GCD(c,d^2st) = 239^2 = d^2st + c$. Clearly $d < 239$ by equations (2). Futhermore, since $GCD(c,d^2st)$ divides $d^2st + c = 239^2$, we have $GCD(c,d^2st) \mid 239^2$, implying $GCD(c,d^2st) \in \{1,239\}$ since 239 is a prime and $GCD(c,d^2st) < 239^2$ by equations (2). Thus we have the following two cases to consider: Case 1: $GCD(c,d^2st)=1$. Then $GCD(s,c)=GCD(t,c)=1$, which inserted in equations (2) result in $(d + 1)^2 = 239^2$, yielding $d=238$, which inserted in equations (2) give us $239^2 = 238^2st + c$, which menns $st < 2$. Therefore $s=t=1$, which inserted in equations (2) result in $c = 239^2 - 238^2 = (239 + 238)(239 - 238) = 477$. Hence $(a,b) = (ds,dt) = (d,d)= (238,238)$, which means there is one solution of equations (1) in this case, namely $(a,b,c) = (238,238,477)$. Case 2: $GCD(c,d^2st) = 239$. Then $GCD(c,st) = 239$ since 239 is a prime and $d<239$. Hence $239 \mid c$ and $239 \mid st$. If $239 \mid s$ and $239 \mid t$, then $239 \mid d^2$ since $(3) \;\; d^2st + c = 239^2$ by equations (1), i.e. $239^2 | c$, which is impossible by equation (3). Hence either $GCD(s,c)=239$ and $GCD(t,c)=1$ or $GCD(s,c)=1$ and $GCD(t,c)=239$, which according to equations (2) yields $d^2 + d + 239d + 239 = 239^2$, or alternatively $(4) \;\; d(d + 240) = 238 \cdot 239$, yielding (since 239 is a prime and $d<239$) $d = 238$, which clearly is not a solution of equation (4). Conclusion: The only solution in positive integers of the equations (1) is $(a,b,c) = (238,238,477)$. q.e.d.