Well, it's easy to show that for any primitive 4D brick (that is, one where a, b, c, and d have no common factor), then exactly one of a, b, c, and d must be odd, and the rest even....
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CSI of Charles University
4d brick
Well, it's easy to show that for any primitive 4D brick (that is, one where a, b, c, and d have no common factor), then exactly one of a, b, c, and d must be odd, and the rest even....