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>> No.7500250 [View]
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7500250

Let's do a raw, straight-up check of the two bricks given in >>7499844 :

<span class="math"> 1008^2 + 1100^2 = 1492^2 \rightarrow 1016064+1210000=2226064 [/spoiler]

<span class="math"> 1008^2 + 1155^2 = 1533^2 \rightarrow 1016064+1334025=2350089 [/spoiler]

<span class="math"> 1100^2 + 1155^2 = 1595^2 \rightarrow 1210000+1334025=2544025 [/spoiler]

<span class="math"> 1008^2 + 1100^2 + 1155^2 = \sqrt{3560089}^2 = g^2 [/spoiler]

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<span class="math"> 1008^2 + 1100^2 = 1492^2 \rightarrow 1016064+1210000=2226064 [/spoiler]

<span class="math"> 1008^2 + 12075^2 = 12117^2 \rightarrow 1016064+145805625=146821689 [/spoiler]

<span class="math"> 1100^2 + 12075^2 = 12125^2 \rightarrow 1210000+145805625=147015625 [/spoiler]

<span class="math"> 1008^2 + 1100^2 + 12075^2 = \sqrt{148031689}^2 = g^2 [/spoiler]

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Each brick has seven components (a,b,c), (d,e,f) and g(^2) , which are the three edge lengths from shortest to longest, the three face diagonals from shortest to longest, and the spatial diagonal. It may happen that c>d or c<d, as these two different bricks themselves demonstrate. In light of the curious irreducibility of each brick's spatial diagonal term, all the components' prime factorizations now follow:

<span class="math"> \displaystyle (1008,1100,1155) \rightarrow (2^4 3^2 7,2^2 5^2 11,3*5*7*11), (2^2 373,3*7*73,5*11*29), 13*17*89*181 [/spoiler]

<span class="math"> \displaystyle (1008,1100,12075) \rightarrow (2^4 3^2 7,2^2 5^2 11,3*5^2*7*23), (2^2 373,3*7*577,5^3*97), 13*29*41*61*157 [/spoiler]

A worthwhile exercise, though nothing too interesting pops out apart from the irreducibility of g^2, and that 12125 term. As a final note on this, note also that the smaller Euler brick (240,252,275) has a spatial diagonal g(^2) with an irreducible prime factorization 13*37*409, just like the other two - that is, there are no even, let alone repeated powers that could come out.

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