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SImplify `(n+1)^2-n^2` Hence form a Pythagorean Triple in which the first integer is 101

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inquire123 | Student | eNoter

Posted June 13, 2013 at 3:18 AM via web

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SImplify `(n+1)^2-n^2`

Hence form a Pythagorean Triple in which the first integer is 101

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted June 13, 2013 at 4:53 AM (Answer #1)

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Simplifying `(n+1)^2-n^2` we get `2^2+2n+1-n^2=2n+1`

Pythagorean triples can be formed by two integers m,n m>n as `m^2-n^2,2mn,m^2+n^2`

If we want 101 to be the first number, then `m^2-n^2=101` . If m=n+1 then 2n+1=101==>n=50 so m=51.

One triple is `51^2-50^2=101,2(51)(50)=5100,51^2+50^2=5101` or

101,5100,5101

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