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I know I need to assume n is odd, then show that there exists an integer k such that n^2 = 8k + 1

so assuming n is odd
i can say n=2j + 1 (where j is an integer)
n^2= (2j +1)^2


but i just can't figure out how to manipulate this in order to prove the conclusion! am i even starting right?

please help! :)

2007-09-26 17:16:13 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

Here is how I did it.

First, you know that every odd integer is 2 greater than the previous positive integer. From that you say for any positive odd integer the value of the next positive odd integer squared is
(n + 2)^2= n^2 + 4n +4

To get the difference between n^2 and (n+2)^2 simply subtract n^2 from n^2 + 4n + 4 =
4n+4

Now replace n with 2j + 1 and you get the difference is
4(2j + 1) + 4 = 8J + 8

Since both 8J and 8 are evenly divisable by 8, their sum is also evenly divisable by 8.

Now if you take the lowest positive odd integer, 1, and show that k exists (It is 0 when n = 1.), by having shown that difference between n^2 and (n+2)^2 is always a multiple of 8 you have proven that k always exists.

2007-09-26 17:55:18 · answer #1 · answered by zman492 7 · 0 0

a million) if A is weird and wonderful then A² is weird and wonderful say, M = ok+a million and N = ok (ok : any constructive integer) M²-N² = (ok+a million)² - ok² = 2k+a million (= any constructive weird and wonderful integer) so there are consistently 2 integers M and N which includes M²-N² : is any weird and wonderful integer examples A = 15 = 2*7+a million ==> ok = 7 15 = 8² - 7² A² = 15² = 225 = 2*112 + a million ==> ok = 112 15² = 113² - 112² 2) if A is even then A² is even say, M = ok+a million and N = ok-a million (ok : any constructive integer) M²-N² = (ok+a million)² - (ok-a million)² = 4k =2²ok A = 2B (B : integer) A² = 2²B² = 2²ok ==> ok = B² to fulfill the undertaking examples A = 18 = 2*9 ==> A² =18² = 2²*9² ==> ok = 80 one 18² = 80 two² - 80² A = 256 = 2*128 ==> A² =256² = 2²*128² ==> ok = 128²=16364 256² = 16365² - 16363² then for any integer A, we are able to discover a minimum of two integers M,N which includes A² = M² - N² ======================================...

2016-11-06 11:54:00 · answer #2 · answered by ? 4 · 0 0

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