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Show that n square -1 is divisible by 8, if n is a positive integer.Explain breifly guys plzz??? |
| Answer» The question is like this:Show that n2\xa0- 1 is divisible by 8, if n is an odd positive integerSoln:Any odd positive integer\xa0n\xa0can be written in form of 4q\xa0+ 1 or 4q\xa0+ 3.\xa0If\xa0n\xa0= 4q\xa0+ 1, then\xa0n2\xa0- 1 = (4q\xa0+ 1)2\xa0- 1 = 16q2\xa0+ 8q\xa0+ 1 - 1 = 8q(2q\xa0+ 1) which is divisible by 8.If\xa0n\xa0= 4q\xa0+ 3,then\xa0n2\xa0- 1 = (4q\xa0+ 3)2\xa0- 1 = 16q2\xa0+ 24q\xa0+ 9 - 1 = 8(2q2\xa0+ 3q\xa0+ 1) which is divisible by 8.\xa0So, n2\xa0- 1 is divisible by 8, if n is an odd positive integer. | |