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Form a Pythagorean triple in which the first integer is 11. Please show your workings...

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phanpal999 | Student, Grade 8 | Salutatorian

Posted June 13, 2013 at 3:16 AM via web

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Form a Pythagorean triple in which the first integer is 11.

Please show your workings clearly.Thanks

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

Posted June 13, 2013 at 4:47 AM (Answer #1)

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Pythagorean triples can be formed using two integers m,n m>n using Euclid's formula: If `a^2+b^2=c^2` then `a=m^2-n^2,b=2mn,c=m^2+n^2`

If 11 is to be the first number, then `m^2-n^2=11`

so we need n such that `sqrt(11+n^2)` is an integer. Trying a few values we find n=5 works: `sqrt(11+5^2)=sqrt(36)=6`

So m=6 and n=5 generating the triple:

`6^2-5^2=11,2(6)(5)=60,6^2+5^2=61`

One triple is 11,60,61

 

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