GCD{$\ \cdot\ $}LCM = PROD for Axles:If {$T\in\mathcal {B}^+$} is an axle, then {$$\text{GCD}_{_T}\ \cdot\ \text{LCM}_{_T} = \text{PROD}_{_T}.$$} Proof: Consider first the case when {$x_{_T}=y_{_T}=1=a_{_T}=b_{_T}=c_{_T}$}, {$T=\langle a,b,c\rangle$}. We have {$\text{GCD}_{_T}=(1,1)$}. The {$1\times 1\times1$} box has no loops, so by {$\Delta$} Respects Loops, {$a\times b\times c$} box has no loops. Thus, the paths on the box completely cover the surface of the box, running through each square twice. That means that {$$4(L^1_{_T}+L^2_{_T}+L^3_{_T})=4\text{PROD}_{_T}.$$} that is, the sum of all the path lengths, {$4(L^1_{_T}+L^2_{_T}+L^3_{_T})$} (there are four specimens of each of the path lengths--The Least Common Multiple of a Box), equals twice the surface area, {$4\text{PROD}_{_T}$} ({$\text{PROD}_{_T}$} is just half the surface area of the box--The Product of a Box). That gives us: {$$(L^1_{_T}+L^2_{_T}+L^3_{_T})=\text{PROD}_{_T},$$} and so {$$\text{GCD}_{_T}\cdot\text{LCM}_{_T}=(1,1)\cdot(L^1_{_T},L^2_{_T}+L^3_{_T})=\text{PROD}_{_T}.$$} If {$x_{_T}=y_{_T}>1$}, then {$a_{_T}=b_{_T}=c_{_T}=x$}, and {$\langle a,b,c\rangle=\langle a'x_{_T},b'x_{_T},c'x_{_T}\rangle$} for some {$a',b',c'$}. The {$\langle a,b,c\rangle$} box, then, is just the expansion of the {$T'=\langle a',b',c'\rangle$}, with all linear dimensions stretched by a factor of {$x_{_T}$}. That gives us {$\text{GCD}_{_{T}}=(x_{_T},x_{_T})=x_{_T}\text{GCD}_{_{T'}}$}, {$\text{LCM}_{_T}=x_{_T}\text{LCM}_{_{T'}}$} and {$\text{PROD}_{_T}=x_{_T}^2\text{PROD}_{_{T'}}$}. Then {$$\text{GCD}_{_T}\cdot\text{LCM}_{_T}=x_{_T}\text{GCD}_{_{T'}}\cdot x_{_T}\text{LCM}_{_{T'}}=x_{_T}^2\text{PROD}_{_{T'}}=\text{PROD}_{_T}.$$} |