We Now Know That C B And C B Are Positive Integers Having No Common Factor That Their Product Is A Square Since C B 1 (68.71 KiB) Viewed 89 times
Number Theory
We Now Know That C B And C B Are Positive Integers Having No Common Factor That Their Product Is A Square Since C B 2 (68.71 KiB) Viewed 89 times
We now know that c-b and c+b are positive integers having no common factor, that their product is a square since (c - b)(c + b) = a2. The only way that this can happen is if c - band c + b are themselves squares. So we can write c+b = 82 and c-b= 1², where s t 1 are odd integers with no common factors. Solving these two equations for band c yields s? + 2 and b= 2 and then a = V(c-b)(c+b) = st. We have (almost) finished our first proof! The following theorem records our accomplishment Theorem 2.1 (Pythagorean Triples Theorem). We will get every primitive Pytha- gorean triple (a, b, c) with a odd and b even by using the formulas 92 + a = st, b = 2 where 8 > t > 1 are chosen to be any odd integers with no common factors. Why did we say that we have "almost" finished the proof? We have shown that if (a,b,c) is a PPT with a odd, then there are odd integers s > t > 1 with no common factors so that a, b, and c are given by the stated formulas. But we still need to check that these formulas always give a PPT. We first use a little bit of algebra to show that the formulas give a Pythagorean triple. Thus (st)? + +2+34 – 2s2t2 + +* _ 5* + 25212 + +* () We also need to check that st, , and have no common factors. This is most easily accomplished using an important property of prime numbers, so we postpone the proof until Chapter 7, where you will finish the argument (Exer- cise 7.3). Thin in intuitiu al ing
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