Editing Pythagorean triple (section)

==Relation to Gaussian integers==

Alternatively, Euclid's formulae can be analyzed and proved using the [[Gaussian integers]].<ref>{{citation |first=John |last=Stillwell|author-link=John Stillwell |title=Elements of Number Theory |chapter-url=https://books.google.com/books?id=LiAlZO2ntKAC&pg=PA110 |year=2002 |publisher=Springer |isbn=978-0-387-95587-2 |pages=110–2 |chapter=6.6 Pythagorean Triples}}</ref> Gaussian integers are [[complex number]]s of the form {{math|''α'' {{=}} ''u'' + ''vi''}}, where {{math|''u''}} and {{math|''v''}} are ordinary [[integer]]s and {{math|''i''}} is the [[imaginary unit|square root of negative one]]. The [[Unit (ring theory)|units]] of Gaussian integers are ±1 and ±i. The ordinary integers are called the [[rational integer]]s and denoted as '{{math|'''Z'''}}'. The Gaussian integers are denoted as {{math|'''Z'''[''i'']}}. The right-hand side of the [[Pythagorean theorem]] may be factored in Gaussian integers:

:<math>c^2 = a^2+b^2 = (a+bi)\overline{(a+bi)} = (a+bi)(a-bi).</math>

A primitive Pythagorean triple is one in which {{math|''a''}} and {{math|''b''}} are [[coprime]], i.e., they share no prime factors in the integers. For such a triple, either {{math|''a''}} or {{math|''b''}} is even, and the other is odd; from this, it follows that {{math|''c''}} is also odd.

The two factors {{math|''z'' :{{=}} ''a'' + ''bi''}} and {{math|''z*'' :{{=}} ''a'' − ''bi''}} of a primitive Pythagorean triple each equal the square of a Gaussian integer. This can be proved using the property that every Gaussian integer can be factored uniquely into Gaussian primes [[up to]] [[unit (ring theory)|units]].<ref name="Gauss_1832">{{Citation | author = Gauss CF | year = 1832 | title = Theoria residuorum biquadraticorum | journal = Comm. Soc. Reg. Sci. Gött. Rec. | volume = 4 | postscript = .| author-link = Carl Friedrich Gauss }} See also ''Werke'', '''2''':67–148.</ref> (This unique factorization follows from the fact that, roughly speaking, a version of the [[Euclidean algorithm]] can be defined on them.) The proof has three steps. First, if {{math|''a''}} and {{math|''b''}} share no prime factors in the integers, then they also share no prime factors in the Gaussian integers. (Assume {{math|1=''a'' = ''gu''}} and {{math|1=''b'' = ''gv''}} with Gaussian integers {{math|''g''}}, {{math|''u''}} and {{math|''v''}} and {{math|''g''}} not a unit. Then {{math|''u''}} and {{math|''v''}} lie on the same line through the origin. All Gaussian integers on such a line are integer multiples of some Gaussian integer {{math|''h''}}. But then the integer ''gh'' ≠ ±1 divides both {{math|''a''}} and {{math|''b''}}.) Second, it follows that {{math|''z''}} and {{math|''z*''}} likewise share no prime factors in the Gaussian integers. For if they did, then their common divisor {{math|''δ''}} would also divide {{math|1=''z'' + ''z*'' = 2''a''}} and {{math|1=''z'' − ''z*'' = 2''ib''}}. Since {{math|''a''}} and {{math|''b''}} are coprime, that implies that {{math|''δ''}} divides&nbsp;{{math|1=2 = (1 + i)(1 − i) = i(1 − i){{sup|2}}}}. From the formula {{math|1=''c''{{sup|2}} = ''zz*''}}, that in turn would imply that {{math|''c''}} is even, contrary to the hypothesis of a primitive Pythagorean triple. Third, since {{math|''c''{{sup|2}}}} is a square, every Gaussian prime in its factorization is doubled, i.e., appears an even number of times. Since {{math|''z''}} and {{math|''z*''}} share no prime factors, this doubling is also true for them. Hence, {{math|''z''}} and {{math|''z*''}} are squares.

Thus, the first factor can be written

:<math>a+bi = \varepsilon\left(m + ni \right)^2, \quad \varepsilon\in\{\pm 1, \pm i\}.</math>

The real and imaginary parts of this equation give the two formulas:

:<math>\begin{cases}\varepsilon = +1, & \quad a = +\left( m^2 - n^2 \right),\quad b = +2mn; \\ \varepsilon = -1, & \quad a = -\left( m^2 - n^2 \right),\quad b = -2mn; \\ \varepsilon = +i, & \quad a = -2mn,\quad b = +\left( m^2 - n^2 \right); \\ \varepsilon = -i, &  \quad a = +2mn,\quad b = -\left( m^2 - n^2 \right).\end{cases}</math>

For any primitive Pythagorean triple, there must be integers {{math|''m''}} and {{math|''n''}} such that these two equations are satisfied. Hence, every Pythagorean triple can be generated from some choice of these integers.

===As perfect square Gaussian integers===
If we consider the square of a Gaussian integer we get the following direct interpretation of Euclid's formula as representing the perfect square of a Gaussian integer.

:<math>(m+ni)^2 = (m^2-n^2)+2mni.</math>

Using the facts that the Gaussian integers are a Euclidean domain and that for a Gaussian integer p <math>|p|^2</math> is always a square it is possible to show that a Pythagorean triple corresponds to the square of a prime Gaussian integer if the hypotenuse is prime.

If the Gaussian integer is not prime then it is the product of two Gaussian integers p and q with <math>|p|^2</math> and <math>|q|^2</math> integers. Since magnitudes multiply in the Gaussian integers, the product must be <math>|p||q|</math>, which when squared to find a Pythagorean triple must be composite. The contrapositive completes the proof.