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Response to pokerfan (Reply #6)

Thu Jan 31, 2013, 07:09 PM

10. Viewed from a sufficiently advanced PoV the Liebniz formula is by itself a proof

Last edited Thu Jan 31, 2013, 09:31 PM - Edit history (1)

that the so-called Gaussian integers a + bi have unique prime factorization, and this unlocks many facts about Pythagorean triples, because |a + bi|^2 = a^2 + b^2

For example, which ordinary primes appear as the hypotenuse of a right triangle? Exactly the ordinary primes that do not remain prime when considered as Gaussian integers: that is, the prime 2 and the primes of the form 4n + 1

For example, if (a,b,c) is a primitive Pythagorean triple, c has no factor of the form 4n + 3. For, consider the equation a^2 + b^2 = (a + bi)(a - bi) = c^2. If we completely factor a + bi in the Gaussian integers and take the complex conjugate of each factor, we get the factorization of a - bi. The prime factorizations of a^2 + b^2 and of c^2 must be the same. If c has an ordinary prime factor q of the form 4n + 3, q does not factor further, so q appears among the factors of a + ib; then q divides each of a, b, c -- contradicting the fact that (a,b,c) is primitive

And so on ...

Again, if (a,b,c) is a primitive Pythagorean triple, consider the (ordinary) prime factorization of c: c = p1*p2*...*pn. No ordinary prime pj here has the form 4n + 3, so each of these ordinary primes factors in the Gaussian integers: pj = (uj + i*vj)(uj - i*vj). Multiplying
(u1 + i*v1)(u2 + i*v2)...(un + i*vn) = (u + i*v)
we find c = (u + i*v)(u - i*v) = u^2 + v^2. So the hypothenuse of a primitive Pythagorean triple is itself a sum of two squares: that, is, a primitive Pythagorean triple is of the form (a,b,u^2 + b^2) and the Pythagorean theorem in this case looks like
a^2 + b^2 = (u^2 + v^2)^2

Using ordinary arithmetic, it is clear that if (a,b,c) is a primitive Pythagorean triple, then a and b cannot both be even; and if they were both odd a^2 + b^2 would be even (but not divisible by 4), so c^2 would be even (but not divisible by 4), so c would be even, so c^2 would be divisible by 4, a contradiction. Therefore one of the legs is even and the other odd. Interchanging a and b if necessary, we may assume a is odd and b is even. Therefore c = u^2 + v^2 = (u + i*v)(u - i*v) is also odd. Since c is odd, the Gaussian prime 1 + i divides neither (u + i*v) nor (u - i*v). Moreover
(a + ib)(a - ib) = a^2 + b^2 = c^2 = (u^2 + v^2)^2
so the Gaussian prime 1 + i divides neither (a + ib) nor (a - ib). But any Gaussian prime that divides both (a + ib) and (a - ib) must be either an ordinary prime of the form 4n + 3 or the Gaussian prime 1 + i, and so (a + ib) and (a - ib) are relatively prime. Each distinct prime factor of c^2 must therefore appear uniquely in (a + ib) and (a - ib); such a prime factor cannot appear in both. So
(a + ib)(a - ib) = c^2 implies both (a + ib) and (a - ib) are squares, say, a + ib = (x + iy)^2 and a - ib = (x - iy)^2
Writing out
a + ib = (x + iy)^2 = x^2 - y^2 + i*2xy
we have a = x^2 - y^2 and b = 2xy
while
(x^2 + y^2)^2 = (x + iy)^2(x - iy)^2 = (a + ib)(a - ib) = a^2 + b^2 = c^2
gives c = x^2 + y^2
So we recover the ancient Greek formula for primitive Pythagorean triples: they all have the form
(x^2 - y^2,2xy,x^2 + y^2)
where exactly one of x and y is even


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Ichingcarpenter Jan 2013 OP
DrDan Jan 2013 #1
intaglio Jan 2013 #2
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Geoff R. Casavant Jan 2013 #4
pokerfan Jan 2013 #5
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struggle4progress Jan 2013 #9
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pokerfan Jan 2013 #6
LineLineNew Reply Viewed from a sufficiently advanced PoV the Liebniz formula is by itself a proof
struggle4progress Jan 2013 #10
struggle4progress Jan 2013 #11
pokerfan Jan 2013 #12
struggle4progress Jan 2013 #13
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