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The Pythagorean Theorem - Educational Outreach

The Pythagorean Theorem - Educational Outreach

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<strong>The</strong> last equality has no solutions that are positive integers.<strong>The</strong>refore, we can end our quest for <strong>Pythagorean</strong> Tripleswhere c b 1.Table 3.2 reveals two ways that Composite<strong>Pythagorean</strong> Triples are formed. <strong>The</strong> first way is when thetwo generators m & n have factors in common, examples ofwhich are m 6 & n 2 , m 6 & n 3 , m 6 & n 4 orm 8 & n 4 . Suppose the two generators m & n have afactor in common which simply means m kp & m kq .<strong>The</strong>n the following is true:a 2mn 2kpkq 2kb mc m22 n n22 ( kp) ( kp)222pq ( kq) ( kq)22 k k22( p( p22 q q22)).<strong>The</strong> last expressions show that the three positive integersa , b & c have the same factor k in common, but now to thesecond power. <strong>The</strong> second way is when the two generatorsm & n differ by a factor of two: m n 2k,k 1,2,3....Exploring this further, we havea 2( n 2k)n 2nb ( n 2k)c ( n 2k)Each of the integers22 n n222 2k 2na b & c 2knn2 4k2 4nk 4n2, is divisible by two. Thus, allthree integers share, as a minimum, the common factortwo.In General: If k 0 is a positive integer and ( a , b,c)is a<strong>Pythagorean</strong> triple, then ( ka , kb,kc)is also a <strong>Pythagorean</strong>Triple. Proof: left as a challenge to the reader..94

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