|
|
| Line 1: |
Line 1: |
| {{about||Lagrange's identity|Lagrange's identity (disambiguation)|Lagrange's theorem|Lagrange's theorem (disambiguation)}}
| | Golda is what's written on my beginning certificate even though it is not the title on my beginning certificate. My working day job is an invoicing officer but I've currently utilized for an additional 1. Ohio is where his house is and his family enjoys it. To play domino is something I truly appreciate doing.<br><br>my weblog - cheap psychic readings - [http://idigdigital.com.au/would-like-to-try-a-whole-new-hobby-here-are-several-suggestions/ http://idigdigital.com.au/would-like-to-try-a-whole-new-hobby-here-are-several-suggestions/] - |
| {{Redirect2|four-square theorem|four square theorem|other uses|four square (disambiguation)}}
| |
| | |
| '''Lagrange's four-square theorem''', also known as '''Bachet's conjecture''', states that any [[natural number]] can be represented as the sum of four integer squares
| |
| | |
| :<math>p = a_0^2 + a_1^2 + a_2^2 + a_3^2\ </math>
| |
| | |
| where the four numbers {{math|''a''<sub>0</sub>, ''a''<sub>1</sub>, ''a''<sub>2</sub>, ''a''<sub>3</sub>}} are integers. For illustration, 3, 31 and 310 can be represented as the sum of four squares as follows:
| |
| | |
| :{{math|3 {{=}} 1<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup> + 0<sup>2</sup>}}
| |
| :{{math|31 {{=}} 5<sup>2</sup> + 2<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup>}}
| |
| :{{math|310 {{=}} 17<sup>2</sup> + 4<sup>2</sup> + 2<sup>2</sup> + 1<sup>2</sup>}}.
| |
| | |
| This theorem was proven by [[Joseph Louis Lagrange]] in 1770, and corresponds to [[Fermat's theorem on sums of two squares]].
| |
| | |
| ==Historical development==
| |
| The theorem appears in the ''[[Arithmetica]]'' of [[Diophantus]], translated into Latin by [[Claude Gaspard Bachet de Méziriac|Bachet]] in 1621.
| |
| | |
| [[Adrien-Marie Legendre]] improved on the theorem in 1798 with his [[Legendre's three-square theorem|three-square theorem]], by stating that a positive integer can be expressed as the sum of three squares if and only if it is not of the form {{math|4<sup>''k''</sup>(8''m'' + 7)}}. His proof was incomplete, leaving a gap which was later filled by [[Carl Friedrich Gauss]].
| |
| | |
| The formula is also linked to [[Descartes' theorem]] of four "kissing circles", which involves the sum of the squares of the curvatures of four circles. This is also linked to [[Apollonian gasket]]s, which were more recently related to the [[Ramanujan–Petersson conjecture]].<ref>[http://www.youtube.com/watch?v=1GtOMVcfjJ0 The Ramanujan Conhecture and some Diophantine Equations], Peter Sarnak, lecture at Tata Institute of Fundamental Research, of the ICTS lecture series. Bangalore, 2013 via youtube.</ref>
| |
| | |
| ==Proof using the Hurwitz integers==
| |
| One of the ways to prove the theorem relies on [[Hurwitz quaternion]]s, which are the analog of [[integer]]s for [[quaternion]]s.<ref name="Stillwell_2003" >{{cite book | author = [[John Stillwell|Stillwell J]] | year = 2003 | title = Elements of Number Theory | publisher = Springer-Verlag | location = New York | isbn = 0-387-95587-9 | pages = 138–157}}</ref> The Hurwitz quaternions consist of all quaternions with integer components and all quaternions with [[half-integer]] components. These two sets can be combined into a single formula
| |
| | |
| :<math>\alpha = \frac{1}{2} E_0 (1 + \mathbf{i} + \mathbf{j} + \mathbf{k}) +E_1\mathbf{i} +E_2\mathbf{j} +E_3\mathbf{k} = a_0 +a_1\mathbf{i} +a_2\mathbf{j} +a_3\mathbf{k}</math>
| |
| | |
| where {{math|''E''<sub>0</sub>, ''E''<sub>1</sub>, ''E''<sub>2</sub>, ''E''<sub>3</sub>}} are integers. Thus, the quaternion components {{math|''a''<sub>0</sub>, ''a''<sub>1</sub>, ''a''<sub>2</sub>, ''a''<sub>3</sub>}} are either all integers or all half-integers, depending on whether {{math|''E''<sub>0</sub>}} is even or odd, respectively. The set of Hurwitz quaternions forms a [[ring (mathematics)|ring]]; that is to say, the sum or product of any two Hurwitz quaternions is likewise a Hurwitz quaternion.
| |
| | |
| The [[field norm|(arithmetic, or field) norm]] {{math|N(''α'')}} of a rational quaternion {{math|''α''}} is the nonnegative [[rational number]]
| |
| | |
| :<math>\mathrm{N}(\alpha) = \alpha\bar\alpha = a_0^2+a_1^2+a_2^2+a_3^2</math>,
| |
| | |
| where <math>\bar\alpha=a_0 -a_1\mathbf{i} -a_2\mathbf{j} -a_3\mathbf{k}</math> is the [[quaternion#Conjugation, the norm, and division|conjugate]] of {{math|''α''}}. Note that the norm of a Hurwitz quaternion is always an integer. (If the coefficients are half-integer, then their squares are a number, equivalent to 1 [[modulo]] 4, divided by 4, so the sum is again an integer.)
| |
| | |
| Since quaternion multiplication is associative, and real numbers commute with other quaternions, the norm of a product of quaternions equals the product of the norms:
| |
| | |
| :<math> \mathrm{N}(\alpha\beta)=\alpha\beta\overline{(\alpha\beta)}=\alpha\beta\overline{\beta}\overline{\alpha}=\alpha \mathrm{N}(\beta)\bar\alpha=\alpha\bar\alpha \mathrm{N}(\beta)= \mathrm{N}(\alpha) \mathrm{N}(\beta).</math>
| |
| | |
| For any <math>\alpha\ne0</math>, <math>\alpha^{-1}=\bar\alpha\mathrm N(\alpha)^{-1}</math>. It follows easily that {{math|''α''}} is a unit in the ring of Hurwitz quaternions if and only if {{math|N(''α'') {{=}} 1.}}
| |
| | |
| The proof of the main theorem begins by reduction to the case of prime numbers. [[Euler's four-square identity]] implies that if the four-square theorem holds for two numbers, it holds for the product of the two numbers. Since any natural number can be factored into powers of primes, it suffices to prove the theorem for prime numbers. It is true for {{math|2 {{=}} 0<sup>2</sup> + 0<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup>}}. To show this for an odd prime integer {{math|''p''}}, represent it as a quaternion {{math|(''p'', 0, 0, 0)}} and assume for now (as we shall show later) that it is not a Hurwitz irreducible; that is, it can be factored into two non-unit Hurwitz quaternions
| |
| | |
| :{{math|''p'' {{=}} ''αβ''}}.
| |
| | |
| The norms of {{math|''p'', ''α'', ''β''}} are integers such that
| |
| | |
| :<math>\mathrm N(p)=p^2=\mathrm N(\alpha\beta)=\mathrm N(\alpha)\mathrm N(\beta)</math>
| |
| | |
| and {{math|N(''α''),N(''β'') > 1.}} This shows that both {{math|N(''α'')}} and {{math|N(''β'')}} are equal to ''p'' (since they are integers), and {{math|''p''}} is the sum of four squares
| |
| | |
| :<math>p=\mathrm N(\alpha)=a_0^2+a_1^2+a_2^2+a_3^2.</math>
| |
| | |
| If it happens that the α chosen has half-integer coefficients, it can be replaced by another Hurwitz quaternion. Choose ω = (±1 ± i ± j ± k)/2 in such a way that γ ≡ ω + α has even integer coefficients. Then
| |
| | |
| :<math>p=(\bar\gamma-\bar\omega)\omega\bar\omega(\gamma-\omega)=(\bar\gamma\omega-1)(\bar\omega\gamma-1).</math>
| |
| | |
| Since γ has even integer coefficients, <math>(\bar\omega\gamma-1)</math> will have integer coefficients and can be used instead of the original α to give a representation of ''p'' as the sum of four squares.
| |
| | |
| As for showing that ''p'' is not a Hurwitz irreducible, [[Joseph Louis Lagrange|Lagrange]] proved that any odd prime {{math|''p''}} divides at least one number of the form {{math|''u'' {{=}} 1 + ''l''<sup>2</sup> + ''m''<sup>2</sup>}}, where {{math|''l''}} and {{math|''m''}} are integers.<ref name="Stillwell_2003" /> This can be seen as follows: since ''p'' is prime, <math>a^2\equiv b^2\pmod p</math> can hold for integers ''a'', ''b'', only when <math>a\equiv\pm b\pmod p</math>. Thus, the set <math>X=\{0^2,1^2,\dots,((p-1)/2)^2\}</math> of squares contains (''p'' + 1)/2 distinct residues modulo ''p''. Likewise, <math>Y=\{-(1+x):x\in X\}</math> contains (''p'' + 1)/2 residues. Since there are only ''p'' residues in total, and <math>|X|+|Y|=p+1>p</math>, the sets ''X'' and ''Y'' must intersect.
| |
|
| |
| The number ''u'' can be factored in Hurwitz quaternions:
| |
| | |
| :{{math|1 + ''l''<sup>2</sup> + ''m''<sup>2</sup> {{=}} (1 + ''l'' '''i''' + ''m'' '''j''')(1 − ''l'' '''i''' − ''m'' '''j''')}}.
| |
| | |
| The norm on Hurwitz quaternions satisfies a form of the [[Euclidean domain|Euclidean]] property: for any quaternion <math> \alpha=a_0+a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k} </math> with rational coefficients we can choose a Hurwitz quaternion <math> \beta=b_0+b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k} </math> so that <math> \mathrm{N}(\alpha-\beta)<1 </math> by first choosing <math> b_0 </math> so that <math> |a_0-b_0| \leq 1/4 </math> and then <math> b_1, b_2, b_3 </math> so that <math> |a_i-b_i| \leq 1/2 </math> for <math> i= </math> 1, 2, 3. Then we obtain
| |
| :<math> \mathrm{N}(\alpha-\beta)=(a_0-b_0)^2+(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2 \leq \left(\frac{1}{4} \right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2=\frac{13}{16}<1. </math>
| |
| It follows that for any Hurwitz quaternions ''α'', ''β'' with ''α'' ≠ 0, there exists a Hurwitz quaternion ''γ'' such that
| |
| :<math>\mathrm N(\beta-\alpha\gamma)<\mathrm N(\alpha).</math>
| |
| The ring H of Hurwitz quaternions is not commutative, hence it is not an actual Euclidean domain, and it does not have [[unique factorization domain|unique factorization]] in the usual sense. Nevertheless, the property above implies that every right ideal is principal. Thus, there is a Hurwitz quaternion ''α'' such that
| |
| :''α''H = ''p''H + (1 − ''l'' '''i''' − ''m'' '''j''')H.
| |
| In particular, ''p'' = ''αβ'' for some Hurwitz quaternion ''β''. If ''β'' were a unit, 1 − ''l'' '''i''' − ''m'' '''j''' would be a multiple of ''p'', however this is impossible as {{math|1/''p'' − ''l''/''p'' '''i''' − ''m''/''p'' '''j'''}} is not a Hurwitz quaternion for ''p'' > 2. Similarly, if ''α'' were a unit, we would have
| |
| :{{math|(1 + ''l'' '''i''' + ''m'' '''j''')H {{=}} (1 + ''l'' '''i''' + ''m'' '''j''')''p''H + (1 + ''l'' '''i''' + ''m'' '''j''')(1 − ''l'' '''i''' − ''m'' '''j''')H ⊆ ''p''H}}
| |
| as ''p'' divides {{math|1 + ''l'' '''i''' + ''m'' '''j'''}}, which again contradicts the fact that {{math|1/''p'' + ''l''/''p'' '''i''' + ''m''/''p'' '''j'''}} is not a Hurwitz quaternion. Thus, ''p'' is not Hurwitz irreducible, as claimed.
| |
| | |
| ==Generalizations==
| |
| Lagrange's four-square theorem is a special case of the [[Fermat polygonal number theorem]] and [[Waring's problem]]. Another possible generalisation is the following problem: Given [[natural number]]s {{math|''a'', ''b'', ''c'', ''d''}}, can we solve
| |
| | |
| :{{math|''n'' {{=}} ''ax''<sub>1</sub><sup>2</sup> + ''bx''<sub>2</sub><sup>2</sup> + ''cx''<sub>3</sub><sup>2</sup> + ''dx''<sub>4</sub><sup>2</sup>}}
| |
| | |
| for all positive integers {{math|''n''}} in integers {{math|''x''<sub>1</sub>, ''x''<sub>2</sub>, ''x''<sub>3</sub>, ''x''<sub>4</sub>}}? The case {{math|1=''a'' = ''b'' = ''c'' = ''d'' = 1}} is answered in the positive by Lagrange's four-square theorem. The general solution was given by [[Ramanujan]]. He proved that if we assume, without loss of generality, that {{math|''a'' ≤ ''b'' ≤ ''c'' ≤ ''d''}} then there are exactly 54 possible choices for {{math|''a'', ''b'', ''c'', ''d''}} such that the problem is solvable in integers {{math|''x''<sub>1</sub>, ''x''<sub>2</sub>, ''x''<sub>3</sub>, ''x''<sub>4</sub>}} for all {{math|''n''}}. (Ramanujan listed a 55th possibility {{math|1=''a'' = 1, ''b'' = 2, ''c'' = 5, ''d'' = 5}}, but in this case the problem is not solvable if {{math|''n'' {{=}} 15}}.<ref>Myung-Hwan Kim [http://icms.kaist.ac.kr/mathnet/kms_tex/974363.pdf REPRESENTATIONS OF BINARY FORMS BY QUINARY QUADRATIC FORMS]</ref>)
| |
| | |
| ==Algorithms==
| |
| | |
| [[Michael O. Rabin]] and [[Jeffrey Shallit]]<ref>M. O. Rabin, J. O. Shallit, ''Randomized Algorithms in Number Theory'', Communications on Pure and Applied Mathematics 39 (1986), no. S1, pp. S239–S256. {{doi|10.1002/cpa.3160390713}}</ref> have found [[randomized algorithm|randomized]] [[polynomial-time algorithm]]s for computing a representation {{math|''n'' {{=}} ''x''<sub>1</sub><sup>2</sup> + ''x''<sub>2</sub><sup>2</sup> + ''x''<sub>3</sub><sup>2</sup> + ''x''<sub>4</sub><sup>2</sup>}} for a given integer {{math|''n''}}, in expected running time {{math|O((log''n'')<sup>2</sup>)}}.
| |
| | |
| ==Uniqueness==
| |
| The sequence of positive integers whose representation as a sum of four squares is unique (up to order) is:
| |
| | |
| :1, 2, 3, 5, 6, 7, 8, 11, 14, 15, 23, 24, 32, 56, 96, 128, 224, 384, 512, 896 ... {{OEIS|A006431}}.
| |
| | |
| These integers consist of the seven odd numbers 1, 3, 5, 7, 11, 15, 23 and all numbers of the form {{math|2 × 4<sup>''k''</sup>, 6 × 4<sup>''k''</sup>}} or {{math|14 × 4<sup>''k''</sup>}}.
| |
| | |
| The sequence of positive integers which cannot be represented as a sum of four ''non-zero'' squares is:
| |
| | |
| :1, 2, 3, 5, 6, 8, 9, 11, 14, 17, 24, 29, 32, 41, 56, 96, 128, 224, 384, 512, 896 ... {{OEIS|A000534}}.
| |
| | |
| These integers consist of the eight odd numbers 1, 3, 5, 9, 11, 17, 29, 41 and all numbers of the form {{math|2 × 4<sup>''n''</sup>, 6 × 4<sup>''n''</sup>}} or {{math|14 × 4<sup>''n''</sup>}}.
| |
| | |
| == See also ==
| |
| * [[Lagrange's three-square theorem]]
| |
| * [[Euler's four-square identity]]
| |
| * [[15 and 290 theorems]]
| |
| * [[Jacobi's four-square theorem]]
| |
| | |
| == Notes ==
| |
| {{reflist}}
| |
| | |
| == References ==
| |
| *{{cite book | author = Ireland and Rosen | title = A Classical Introduction to Modern Number Theory | publisher = Springer-Verlag | year = 1990 | isbn=0-387-97329-X }}
| |
| | |
| == External links ==
| |
| * [http://planetmath.org/encyclopedia/ProofOfLagrangesFourSquareTheorem.html Proof at PlanetMath.org]
| |
| * [http://www.alpertron.com.ar/4SQUARES.HTM Another proof]
| |
| * [http://www.alpertron.com.ar/FSQUARES.HTM an applet decomposing numbers as sums of four squares]
| |
| | |
| {{DEFAULTSORT:Lagrange's Four-Square Theorem}}
| |
| [[Category:Additive number theory]]
| |
| [[Category:Theorems in number theory]]
| |
| [[Category:Articles containing proofs]]
| |