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Form a Pythagorean triple in which the first integer is 11. Please show your workings...
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High School Teacher
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:
One triple is 11,60,61
Posted by embizze on June 13, 2013 at 4:47 AM (Answer #1)
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