If it had one solution $(a_{0},b_{0},c_{0})$, then for any $p,q,r$, suth that $(p,qra_{0}b_{0}c_{0})=1=(q,pra_{0}b_{0}c_{0})=(r,qpa_{0}b_{0}c_{0})$ $(a_{0}p,b_{0}q,c_{0}r)$ is solution too.

I resolved the problem later, but I do not know the origin of it Can anyone help me? Solve the equation : $ ab+c= \ (a^{2},b^{2})+(a,bc)+(b,ac)+(c,ab)=239^2$ in positive integers.

math10 wrote: I resolved the problem later, but I do not know the origin of it Can anyone help me? Solve the equation : $ ab + c = \ (a^{2},b^{2}) + (a,bc) + (b,ac) + (c,ab) = 239^2$ in positive integers. We have $ (a,b,c)|\ (a^{2},b^{2}) + (a,bc) + (b,ac) + (c,ab) =239^2$ If $ 239|(a,b,c)$, then $ ab+c \ge (239)(239)+239>239^2$, contradiction.So $ (a,b,c)=1$. So let $ (a,b)=x,(b,c)=y,(c,a)=z$ the equation becomes : $ ab+c = x^2+zx+xy+yz=239^2$ with $ (zx)(xy)|ab$ and $ (yz)|c$ So $ yz|(ab+c)=239^2$. As $ (y,z)=1$ and that $ 239$ is a prime, we have at least one of $ y,z$, say $ y=1$ $ \Leftrightarrow ab+c=(x+1)(x+z)=239^2 \Rightarrow \begin{cases}x+1=239 \\ x+z=239 \end{cases} \Rightarrow \begin{cases}x=238 \\ z=1 \end{cases}$ As $ x^2|ab$, we have $ x^2=ab$ or $ ab+c >2 (238)^2>239$, contradiction. So $ a,b=x=238$, $ c=239^2-238^2=477$ is the solution after checking.

Hint

my solutions of problem is to prove $ a = b$. If $ a,b$ are different.We have: $ ab > (a;b)^2 + a + b$. But:$ (a;b)^2 + (a;bc) + (b;ca) + (c;ab)\leq(a;b)^2 + (a + b + c)$ So $ ab\leq(a;b)^2 + a + b$. $ \Rightarrow a=b$

mathmanman wrote: Solve the equation \[ (a^{2},b^{2}) + (a,bc) + (b,ac) + (c,ab) = 199. \] in positive integers. (Here $ (x,y)$ denotes the greatest common divisor of $ x$ and $ y$.) Firste note that $ 199$ is a prime. Assume that $ (a,b,c) = d > 1$. Then $ d^2 \mid (a^2,b^2)$ og $ d \mid (a,bc), (b,ac), (c,ab)$. Then since $ d >1, d \mid 199$ we see that $ d = 199$. And hence $ LHS \ge 199^2 + 3 \cdot 199 > 199$. Contradiction. So $ (a,b,c)=1$. We can see that $ (a,bc) = (a,b) \cdot (a,c)$. Proof: Let $ (b,c) = d$, and $ b = db_1, c = dc_1$ (then $ (b_1,c_1)=1$). Since $ (a,b,c)=1$ we see that $ (a,d)=1$ hence $ (a,bc) = (a,b_1c_1) = (a,b_1) \cdot (a,c_1) = (a,b) \cdot (a,c)$. It's obvious that $ (a^2,b^2) = (a,b)^2$. So $ LHS = (a,b)^2 + (a,b)\cdot (a,c) + (a,b) \cdot (b,c) + (a,c) \cdot (b,c) =$ $ ( (a,b) + (a,c) ) \cdot ( (a,b) + (b,c) ) = 199$ Since $ 199$ is a prime, this can't happen. (Of course $ (a,b), (a,c), (b,c) \ge 1$) So: No solutions.

nice solution. thank Mathias_DK very much.