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How to find the all positive integer triples such that :

$$ab+c=\gcd (a^2,b^2)+\gcd(a,bc)+\gcd(b,ac)+\gcd(c,ab)=239^2$$

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up vote 2 down vote accepted

Let $x=\gcd(a,b,c)$. But $\gcd(a^2,b^2)\ge x^2$ and $x$ divides $239^2$, this is impossible due to the equality and 239 is prime. Hence $x=1$.

Let $y=\gcd(c,ab)$. if $y>1$, then once again $y=239$ ($239^2$ would be too big). Suppose $y$ divides $a$ (the other case is symmetric), $y$ does not divide $b$ ($\gcd(a^2,b^2)$ would be at least $239^2$, too big). Hence $\gcd(c,ab)=y$, $\gcd(a,bc)=y$, and $\gcd(a,bc)=\gcd(a,b)=z$, so $$ z^2+z+2y=y^2$$ So $y$ divides $z$ (or $z+1$), this is not possible, so $y=1$

So $c$ is prime with $a$ and $b$ and the equation is $$ab+c=\gcd(a,b)^2+2\gcd(a,b)+1=(\gcd(a,b)+1)^2=239^2$$ $$\gcd(a,b)=238$$

The only small enough possibility is $a=b=238$ and then $c=477$

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