Prove that the sum of the squares of two consecutive odd integers leaves remainder 2 on division by 8

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05-11-2017
Mathematics

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Prove that the sum of the squares of two consecutive odd integers leaves remainder 2 on division by 8

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Rod44 Rod44 · Answer 1 · 2017-11-05T20:34:59+01:00

Rod44 Rod44

05-11-2017

Two consecutive odd integers can be represented by 2x-1 and 2x+1 where x is any integer.
The sum of the squares is 4x²-4x+1+4x²+4x+1=8x²+2. If we divide by 8 we get x² and a remainder of 2.

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konrad509 konrad509 · Answer 2 · 2019-06-19T22:32:42+02:00

konrad509 konrad509

19-06-2019

[tex](2n+1)^2+(2n+3)^2=4n^2+4n+1+4n^2+12n+9=\\=8n^2+16n+10=8n^2+16n+8+2=8(n^2+2n+1)+2[/tex]

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