GCD and LCM 2

Let $gcd(a,b,c,d) = S$ ,

$gcd(a,b,c) = S S_{abc}$ , $gcd(a,b,d) = S S_{abd}, ... etc$ ,

$gcd(a,b) = S S_{abc} S_{abd} S_{ab}$ , $gcd(a,c) = S S_{abc} S_{acd} S_{ac}, ... etc$ ,

$a = S S_{abc} S_{abd} S_{acd} S_{ab} S_{ac} S_{ad} A, b = S S_{abc} S_{abd} S_{bcd} S_{ab} S_{bc} S_{bd} B, ... etc$ ,

$S_3 = S_{abc} S_{abd} S_{acd} S_{bcd}$ ,

$S_2 = S_{ab} S_{ac} ... S_{cd}$ ,

$S_1 = ABCD$ .

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$lcm(a,b,c,d) = S S_3 S_2 S_1.$

$lcm(a,b,c) = \frac{lcm(a,b,c,d)}{D}.$

$lcm(a,b) = \frac{lcm(a,b,c,d)}{S_{cd} C D}.$

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$\frac{gcd(a,b,c,d)^{x_4} }{ (\prod_{i,j \in (a,b,c,d); i < j} gcd(i,j))^{x_2}(\prod_{i,j,k \in (a,b,c,d); i < j < k} gcd(i,j,k))^{x_3} } = \frac{S^{x_4} }{(S^6 S_3^3 S_2)^{x_2} (S^4 S_3)^{x_3} },$

$\frac{(\prod_{i,j,k \in (a,b,c,d); i < j < k} lcm(i,j,k))^{y_3} }{lcm(a,b,c,d)^{y_4} (\prod_{i,j \in (a,b,c,d); i < j} lcm(i,j))^{y_2}} = \frac{(S^4 S_3^4 S_2^4 S_1^3)^{y_3} }{ (S S_3 S_2 S_1)^{y_4} (S^6 S_3^6 S_2^5 S_1^3)^{y_2} }.$

Comparing orders of:

$S_1: 0 = 3y_3 - (y_4 + 3y_2)$ , so $y_4 = 3(y_3 - y_2)$

$S_2: x_2 = -4y_3 + (y_4 + 5y_2) = -y_3 + 2y_2$

$S_3: -3x_2 - x_3 = 4y_3 - (y_4 + 6y_2)$ , so $x_3 = -3x_2 - 4y_3 + (y_4 + 6y_2) = 2y_3 - 3y_2$

$S: x_4 - (6x_2 + 4x_3) = 4y_3 - (y_4 + 6y_2)$ , so $x_4 = 6x_2 + 4x_3 + 4y_3 - (y_4 + 6y_2) = 3(y_3 - y_2)$ .

These are all positive integers, so $0 < 3y_2 < 2y_3 < 4y_2$ .

The sum $x_2 + x_3 + x_4 + y_2 + y_3 + y_4 = (y_3 - 2y_2) + (2y_3 - 3y_2) + (3y_3 + 3y_2) + y_2 + y_3 + (3y_3 - 3y_2) = 8y_3 - 6y_2$ .

Note, $4y_3 < 8y_3 - 6y_2 < 5y_3$ .

The smallest value of $y_3$ that fits in $(3y_2, 4y_2)$ for some $y_2$ , is $y_3 = 5$ , with $y_2 = 3$ .

$8y_3 - 6y_2 = 22$ in this case. Note, for larger values of $y_3$ , $8y_3 - 6y_2 > 24 > 22$ .

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Hence, $22$ is the minimum.

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(In this case, $x_2 = 1, x_3 = 1, x_4 = 6, y_2 = 3, y_3 = 5, y_4 = 6$ ).