mihaizubcopb6vp4 mihaizubcopb6vp4
  • 01-07-2018
  • Mathematics
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Prove algebraically that the square of any odd number is always 1 more that a multiple of 8.

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konrad509
konrad509 konrad509
  • 01-07-2018
[tex](2n+1)^2=4n^2+4n+1=4n(n+1)+1[/tex]

[tex]n(n+1)[/tex] is a product of two consecutive numbers, so it's divisible by 2. Therefore, the product [tex]4n(n+1)[/tex] is divisible by [tex]4\cdot2=8[/tex]. In other words, that product is a multiple of 8. So [tex]4n(n+1)+1[/tex] is always "1 more that a multiple of 8".


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