MHB Determine positive integer solutions

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The equation x + y + z + xy + yz + xz = xyz + 1 seeks positive integer solutions for triples (x, y, z) under the condition z ≥ y ≥ x. Various methods, including substitution and analysis of small values, are discussed to find valid triples. The importance of symmetry in the variables is highlighted, as it can simplify the search for solutions. The discussion emphasizes the need for systematic approaches to ensure all possible combinations are considered. Ultimately, the goal is to identify all valid triples that satisfy the given equation.
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Determine all triples of positive integers $x,\,y,\,z$ such that $z\ge y\ge x$ and $x+y+z+xy+yz+xz=xyz+1$.
 
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anemone said:
Determine all triples of positive integers $x,\,y,\,z$ such that $z\ge y\ge x$ and $x+y+z+xy+yz+xz=xyz+1$.

add $1 + xyz$ on both sides to get $(1+x)(1+y)(1+z) = 2(1+xyz)$

for upper bound on x letting $x =y = z$ we get $(1+x)^3 =2(1+x^3)=> x <4$

so we need to check for 3 cases $x = 1, 2 , 3$

$x=1 => 2(1+y)(1+z) = 2(1+yz) => 1+y+z + yz = 1 + yz=> y+ z = 0$ or no solution$x=2=> 3(1+y)(1+z) = 2(1+2yz)$
or $3 + 3y + 3z+ 3yz = 2 + 4yz$
or $ yz-3y-3z =1$
or $(y-3)(z-3) = 10$
giving one solution $z-3=10,y-3=1=> z = 13,y=4$
or another solution $z-3 = 5,y-3 =2 => z= 8,y = 5$

$x=3=> 4(1+y)(1+z) = 2(1+3yz)$
or $4 + 4y + 4z + 4yz = 2 + 6yz$
or $ yz - 2y - 2z = 1$
or $(y-2)(z-2) = 5$ giving $z-2=5, y-2 = 1$ or $z = 7, y = 3$

so 3 sets of solution $(x,y,z) = (2,5,8),(2,4,13),(3,3,7)$
 
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Well done kaliprasad! :cool:
 
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