The Triangular Generic Claim:If {$T\in\mathcal {B}^+$} is triangular and generic, then {$$\text{GCD}_{_T}\cdot\text{LCM}_{_T}=\text{PROD}_{_T}.$$} Proof: If {$T$} is triangular, then {$T=\langle a_{_T},b_{_T},c_{_T}\rangle)$} and {$\Delta(T)=\langle a_{_T},b_{_T},c_{_T}-a_{_T}-b_{_T}\rangle$}, {$x_{_T}=a_{_T}+b_{_T}-c_{_T}$}. When we apply {$\Delta$} successively to {$\langle a_{_T},b_{_T},-x_{_T}\rangle$}, we perform exactly the same steps as applying the Euclidean algorithm to {$a_{_T}-x_{_T},b_{_T}-x_{_T}$}. As before, we are using a ``slow'' Euclidean algorithm, where we successively subtract the smaller number from the larger. The first step here, since {$b_{_T}\ge a_{_T}$}, is {$\Delta(\langle a_{_T},b_{_T},-x_{_T}\rangle)=\langle a_{_T}-(b_{_T}-x_{_T}),b_{_T},-x_{_T}\rangle$}. Here's an example, {$\langle 11,15,-5\rangle$}, with the Euclidean algorithm applied to {$(11-5,15-5)=(6,10)$}:
In the end we arrive at {$\langle x_{_T},y_{_T},-x_{_T}\rangle$}, where {$y_{_T}$} is the greatest common divisor of {$a_{_T}-x_{_T},b_{_T}-x_{_T}$} plus {$x_{_T}$}, that is, {$y_{_T}=(a_{_T}-x_{_T},b_{_T}-x_{_T})+x_{_T}=(c_{_T}-b_{_T},c_{_T}-a_{_T})+x_{_T}$} (using {$x_{_T}=a_{_T}+b_{_T}-c_{_T}$}). This gives us that {$z_{_T}=(c_{_T}-b_{_T},c_{_T}-a_{_T})$}, so {$$\text{GCD}_{_T}=(a_{_T}+b_{_T}-c_{_T},(c_{_T}-b_{_T},c_{_T}-a_{_T})).$$} From Just Right and the fact that {$T$} is not an axle or a spoke, {$L_1=L_2=a_{_T}+b_{_T}+c_{_T}$}. Computing {$L_3$}, however, is a delicate task. Length Computation Subclaim: {$$L_3=I+\left(\frac{c_{_T}-a_{_T}}d+ \frac{c_{_T}-b_{_T}}d-1\right)(a_{_T}+b_{_T}-c_{_T}) +\left(\frac{c_{_T}-a_{_T}}d+\frac{c_{_T}-b_{_T}}d-2\right)c_{_T},$$} where {$I$} is the least common multiple of {$c_{_T}-b_{_T}$} and {$c_{_T}-a_{_T}$} and {$d$} is the greatest common divisor. Now we compute {$\text{GCD}_{_T}\cdot\text{LCM}_{_T}$} using {$I\cdot d=(c_{_T}-b_{_T})(c_{_T}-a_{_T})$} and writing {$n$} for {$\frac{c_{_T}-a_{_T}}d+\frac{c_{_T}-b_{_T}}d-1$}: {$$\begin{array}{rcl}&&\text{GCD}_{_T}\cdot\text{LCM}_{_T}\\ &=&(a_{_T}+b_{_T}-c_{_T},d)\cdot(a_{_T}+b_{_T}+c_{_T},a_{_T}+b_{_T}+c_{_T}+ \frac{(c_{_T}-b_{_T})(c_{_T}-a_{_T})}d+n(a_{_T}+b_{_T}-c_{_T})+(n-1)c_{_T})\\ &=&(a_{_T}+b_{_T})^2-{c_{_T}}^2+d(a_{_T}+b_{_T}+c_{_T})+(c_{_T}-b_{_T})(c_{_T}-a_{_T})+dn(a_{_T}+b_{_T}-c_{_T}) +d(n-1)c_{_T}\\ &=&\text{(eventually) }a_{_T} b_{_T}+a_{_T} c_{_T}+ b_{_T} c_{_T}. \end{array}$$} This verifies the proposition for {$T=\langle a_{_T},b_{_T},c_{_T}\rangle$}. |
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