Visitor

2015-12-14 03:24:38

#1

Show that 100^{2013} could be written as a sum of perfect squares.

Member

2015-12-14 03:39:40

#2

100^{2013} = 100·100^{2012} = (64 + 36)·100^{2012} = (8^{2} + 6^{2}) · (100^{1006})^{2} = 8^{2} · (100^{1006})^{2} + 6^{2} · (100^{1006})^{2} = (8·100^{1006})^{2} + (6·100^{1006})^{2}

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