Problem 855

Solution: Rewrite the system

$$\left\{\begin{array} { l } { a ( x + y ) = b ( z - x ) ; } \\ { a ( x ^ { 3 } + y ^ { 3 } ) = b ( z ^ { 3 } - x ^ { 3 } ) ; } \end{array} \Longrightarrow \left\{\begin{array}{l} a(x+y)=b(z-x) \\ a(x+y)\left(x^{2}-x y+y^{2}\right)=b(z-x)\left(z^{2}+z x+x^{2}\right) \end{array}\right.\right.$$

Since $x, y, z$ are natural numbers, then $a(x+y)=b(z-x)>0$. Therefore, in the second equation, we can divide both sides by the same non-zero number.

$$\left\{\begin{array} { l } { a ( x + y ) = b ( z - x ) ; } \\ { x ^ { 2 } - x y + y ^ { 2 } = z ^ { 2 } + z x + x ^ { 2 } ; } \end{array} \Longrightarrow \left\{\begin{array} { l } { a ( x + y ) = b ( z - x ) ; } \\ { ( y - z ) ( y + z ) = x ( y + z ) ; } \end{array} \Longrightarrow \left\{\begin{array}{l} a(x+y)=b(z-x) \\ y-z=x \end{array}\right.\right.\right.$$

Substitute $x$ with $y-z$ in the first equation

$$\left\{\begin{array} { l } { a ( 2 y - z ) = b ( 2 z - y ) ; } \\ { x = y - z } \end{array} \Longrightarrow \left\{\begin{array}{l} (2 a+b) y=(a+2 b) z \\ x=y-z \end{array}\right.\right.$$

Since $y$ and $z$ are coprime, then $y=a+2 b$ and $z=2 a+b$. Therefore, $x=y-z=(a+2 b)-(2 a-b)=b-a$. Since $b>a$ are coprime natural numbers, $x, y, z$ are pairwise coprime numbers.

It remains to compute

$$x+y+z=(b-a)+(a+2 b)+(2 a+b)=2 a+4 b$$

## Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
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| Answer | 34 | 44 | 38 | 40 | 38 | 58 | 46 | 46 | 50 | 40 | 46 |