The Ramanujan-Nagell Theorem: Understandng the Proof By Spencer De Chenne 1 Introducton The Ramanujan-Nagell Theorem, frst proposed as a conjecture by Srnvasa Ramanujan n 1943 and later proven by Trygve Nagell n 1948, largely owes ts proof to Algebrac Number Theory (ANT) As the reader mght have taken from the name, ANT expands and often reles on results from ordnary Number Theory; however, as ANT should be read as the theory of Algebrac Numbers, t s natural to thnk of ANT as an extenson of Abstract Algebra, and as such the reader should be famlar wth many results from Abstract Algebra, and especally confdent wth ther basc understandng of Felds More specfcally, whle Number Theory focuses largely on the study of the ntegers, N, ANT focuses on the ratonal numbers, Q, and extensons of Q In ths manner, we shall begn by showng some prelmnary results and defntons Algebrac Number Theory Essentals Theorem 1: The set Q of all algebrac numbers over Q s a subfeld of C Proof We wll use the result from Abstract Algebra that for α E, for a felds F and E F (α), α s algebrac f and only f [E : F ] s fnte Now suppose that α and β are algebrac over Q Then [Q(α, β) : Q] [Q(α, β) : Q(α)][Q(α) : Q] Now, snce β s algebrac over Q, t must certanly be algebrac over Q(α) and the extenson must be fnte, and snce α s algebrac over Q ths extenson must also be fnte Therefore, [Q(α, β) : Q] s fnte and Q(α, β) must be a fnte extenson over Q Furthermore, α + β, α β, αβ, and α/β (for β 0) must belong to Q(α, β) Thus, n ths way, adjonng algebrac elements wll always result n fnte extensons, and Q must be a subfeld of C Whle Q may not be very nterestng n ts entrety (due perhaps to the fact that [Q : Q] s not fnte), we are very nterested n number felds, whch are subfelds K of C such that [K : Q] s fnte Ths mples that every element of K s algebrac, and hence K s a subfeld of Q Just as Abstract Algebra generalzes many concepts the reader learned n hgh school, ANT generalzes many results and concepts from ordnary Number Theory One such nterestng and mportant concept s a generalzaton of the ntegers, known as algebrac ntegers Smlar to the defnton of algebrac numbers, an algebrac nteger s a complex number α such that α s the root of a monc polynomal p(x) Z[x]; n other words, p(α) n k α 0 0 1
for k Z, k n 1 It s easy to observe that not every element of Q s an algebrac nteger Take, as an example, φ /7, whch s a root of the polynomal q(x) 7x, whch s not monc, and s(x) x /7 / Z[x] Further studes shall be conducted n fnte extensons of Q, and often we are concerned wth the algebrac ntegers of such extenson felds For ths reason, we shall show some prelmnary results about the algebrac ntegers, denoted A Theorem : The algebrac ntegers form a subrng of the feld of algebrac numbers Proof We must show that for α, β A, α + β A and αβ A Suffce t to say that for φ C, φ s algebrac f and only f the addtve group generated by all powers 1, φ, φ, s fntely generated From ths we can clam that the powers of α + β and αβ le n a fntely generated subgroup of C, and so α + β and αβ are algebrac Hence, A forms a subrng of Q Theorem 3: An algebrac number α s an algebrac nteger f and only f ts mnmum polynomal over Q has coeffcents n Z Proof Let p(x) be the mnmum polynomal of α over Q, and recall that ths s both monc and rreducble n Q[x] If p(x) Z[x], then α f an algebrac nteger Conversely, f α s algebrac, then q(α) 0 for some q(x) Z[x], and p q It follows that p Z[x] because some ratonal multple γp Z[x] and dvdes q, and the moncty of q mples γ 1 Smlar to Q, we are rarely nterested n the entrety of A; however, for fnte extensons of Q, we are often nterested n the algebrac ntegers contaned n the feld For an extenson feld K over Q, we defne the rng of ntegers of K as O K K A Snce both A and K are subrngs of C, then t follows that O K s a subrng of K Furthermore, because Z Q K and Z A, then Z O K Wth the concept of algebrac ntegers, we can redefne modular arthmetc n a natural way We say that for α, β, γ O K, f α β (mod γ), then α γφ + β for some φ O K To demonstrate ths concept, we shall use an example that wll llumnate methods of verfcaton as well as serve us n a future proof Example: For a, b Q( 7), consder a 1 + 7 and b 1 7 It s smple to verfy that a and b are both algebrac ntegers (as well as conjugates) because they both satsfy the polynomal p(x) x + 3x + 4 We clam that To verfy ths, consder the quotent a 1 (mod b ) a 1 b 1 7
We leave t to the reader to verfy that the quotent s a root of x x +, a monc polynomal wth coeffcents n Z Now that we have defned what modular arthmetc wth algebrac ntegers s, we can state the followng theorem that wll be useful n understandng the Ramanujan-Nagell Theorem Keep n mnd that a squarefree ratonal nteger s an element a Z such that the prme factorzaton of a has no squared factors Theorem 4: Let d be a squarefree ratonal nteger Then the ntegers of Q( d) are: [ ] 1 Z d f d 1 (mod 4), [ 1 Z + 1 ] d f d 1 (mod 4) Proof Every element α Q( d) s of the form α r + s d, for r, s Q Hence, we may wrte α a + b d c where a, b, c Z, c > 0, and no prmes dvde every a, b, and c Now α s an nteger f and only f the coeffcents of the mnmum polynomal ( ( a + b )) ( ( d a b )) d x x c c are ntegers Thus, a b d Z, (1) c a Z () c If c and a have a common prme factor p then (1) mples that p dvdes b (snce d s squarefree) whch contradcts our prevous assumpton Hence, from () we have c 1 or If c 1, then α s an nteger of Q( d) n any case, so we may concentrate on the case c Now a and b must both be odd, and (a b d)/4 Z Hence, a b d 0 (mod 4) Now, and odd number k + 1 has square k + 4k + 1 1 (mod 4), hence a 1 b (mod 4), and ths mples d 1 (mod 4) Conversely, f d 1 (mod 4) then for odd a, b we have α an nteger because (1) and () hold To sum up: f d 1 (mod 4), then c 1 and so (1) holds; whereas f d 1 (mod 4) we can also have c and a and b odd, whence easly () holds The reader may be nterested to know that such felds are called quadratc felds f they are a degree extenson over Q In the case where d s postve, such felds are known as real felds, 3
whereas f d s negatve the felds are known as magnary felds Currently, as the reader may have guessed, we are nterested n the latter Consder another example of the magnary quadratc feld Q( 7): Example: 7 1 (mod 4), whch mples that the ntegers of Q( 7) are Z [ 1 + 1 7 ] In order to prove some upcomng theorems, the reader should be famlar wth some results from feld theory The followng theorem should seem both famlar and new Theorem 5: Let K Q(α) be a number feld of degree n over Q Then there are exactly n dstnct monomorphsms σ : K C ( 1,, n) whch fx Q element-wse The elements σ (α) α are the dstnct zeros n C of the mnmum polynomal of α over Q Proof It s an mportant and non-trval result that an rreducble polynomal over a subfeld K of C has no repeated roots n C; however, ths result wll not be proved here Let α 1,, α n be the n dstnct roots of the mnmum polynomal p(x) of α guaranteed by ths result Then each α also has mnmum polynomal p(x), and so there s a unque feld somorphsm σ : Q(α) Q(α ) such that σ (α) α In fact, f β Q(α) then β r(α) for a unque r Q[x] wth deg(r) < n; and we must have σ (β) r(α ) Conversely, f σ : K C s a monomorphsm then σ s the dentty on Q Then we have 0 σ(p(α)) p(σ(α)) so that σ(α) s one of the α, hence σ s one of the σ Example: Consder the feld Q( 7) Because ths s a degree extenson over Q and both roots of the mnmal polynomal of 7 are contaned n Q( 7), there must be exactly two automorphsms that fx Q element-wse The automorphsms are then: σ 1 (a + b 7) a + b 7 σ (a + b 7) a b 7 Ths specfc result, resemblng prevous studes n abstract algebra, ads us n the followng defnton: Defnton: Let K Q(α) be a degree n extenson The norm of β K s N K (β) n σ (β), 1 where the σ are the monomorphsms prevously defned Example: Consder agan K Q( 7), and let α a + b 7 K Then N K (α) (a + b 7)(a b 7) a 7b 4
At ths pont, we leave t to the reader to remnd themselves of the defntons of prmes, unts and Noetheran rngs, as these are mportant concepts n the followng secton The reader should be aware that for an ntegral doman D, factorzaton nto rreducbles s possble f D s Noetheran Furthermore, for a number feld K, O K s Noetheran (the proof wll not be gven here, as t requres more theory of free groups than s expected of the reader) From ths, t s a drect corollary to state that factorzaton nto rreducbles s possble n O K The followng theorem dscusses unts n specfc felds: Theorem 6: The group of unts U of the ntegers n Q( d) where d s negatve and squarefree s as follows: 1 For d 1, U {±1, ±} For d 3, U { ±1, ±ω, ±ω } where ω e π/3 3 For all other d < 0, U {±1} The proof of ths theorem s too broad to be added to ths paper For clarfcaton, the reader can refer to Stewart and Tall s Algebrac Number Theory As one mght have gathered, our prmary nterest s n the feld Q( 7) The mplcaton of ths theorem s that we can now confdently state that the group of unts of ths feld s {±1} The followng theorem tes together many prevous deas and s perhaps the most powerful n our proof of the Ramanujan-Nagell theorem, as t addresses factorzaton n number felds Theorem 7: Let O K be the rng of ntegers of a number feld K, and let x, y O K Then 1 x s a unt f and only f N(x) ±1, If x and y are assocates, then N(x) ±N(y), 3 If N(x) s a ratonal prme, then x s rreducble n O K Proof 1) f xu 1, then N(x)N(u) 1 Snce N(x), N(u) Z (see Stewart and Tall s Algebrac Number Theory), we have N(x) ±1, then σ 1 (x)σ (x)σ n (x) ±1 whch s the defnton of the norm of x One factor, wthout loss of generalty σ 1 (x), s equal to x (from the dentty mappng); all other σ (x) are ntegers Put u ±σ (x)σ n (x) Then xu 1, so u x 1 K Hence u K A O, and x s a unt ) If x, y are assocates, then x uy for a unt u, so N(x) N(uy) N(u)N(y) ±N(y) by 1) 3) Let x yz Then N(y)N(z) N(yz) N(x) p, a ratonal prme; so one of N(y) and N(z) s ±p and the other s ±1 By 1), one of y and z s a unt, so x s rreducble The followng theorem wll not be proven here due to both the proof s length and complexty; however, we are nterested n only one of the results from the theorem, whch wll valdate many of the conclusons drawn n the Ramanujan-Nagell Theorem 5
Theorem 8: The rng of ntegers O of Q( 7) s Eucldan for d 1,, 3, 7, 11, wth Eucldan valuaton φ(α) N(α) Proof See Stewart and Tall s Algebrac Number Theory At ths pont, the reader should revew examples prevously stated n the text, as every sngle one was created n order to ease some of the understandng of the followng secton As one could gather, the followng secton focuses on applcatons of the learned materal n the feld Q( 7) 3 The Ramanujan-Nagell Theorem Questons n Algebrac Number Theory often seem at frst glance to be questons n ordnary Number Theory The Ramanujan-Nagell Theorem s such a theorem, whose concluson s about the nteger solutons to an equaton However, we shall see that the proof utlzes feld extensons and propertes of unque factorzaton n order to state that the presented solutons are the only solutons to the equaton Theorem (Ramanujan-Nagell): The only solutons to the equaton for x, n Z are x + 7 n ±x 1 3 5 11 81 n 3 4 5 7 15 Proof We wll work n Q( 7), whose rng of ntegers has unque factorzaton Clearly, x must be odd, and we see that for (x, n) whch satsfes the equaton, ( x, n) too must satsfy the equaton, and so we wll assume x to be postve Frst, we assume that n s even, n whch case we have the factorzaton 7 n x ( n/ x)( n/ + x) Clearly, both n/ x and n/ + x must be ntegers Because x s assumed postve and n > 0, then n/ + x > n/ x, and we fnd that 7 n/ + x 1 n/ x, from whch we observe 8 1+n/ 6
Thus, n 4, and we fnd that x 3 Now let n be odd, and assume n > 3 We can see that ( ) ( ) 1 + 7 1 7 s a factorzaton nto prmes Obvously, x must be odd, so let x k + 1, mplyng that x + 7 4k + 4k + 8 s dvsble by 4 We can substtute m n and rewrte the orgnal equaton to be solved as x + 7 m, 4 so that ( ) ( ) x + 7 x 7 1 + 7 1 7, where the rght-hand sde s a prme factorzaton Nether 1+ 7 nor 1 7 s a common factor of the left-hand sde, because such a factor would dvde ther dfference, 7, whch s seen to be mpossble by takng ther norms Comparng the two factorzatons, snce the only unts of Q( 7) are ±1 (whch we showed prevously), we must have x ± 7 1 ± 7 ± From ths, we can see that x + 7 x 7 x + 7 x 7 1 + 7, 1 7, 1 7, 1 + 7 By takng dfferences of these equatons, we derve that ± 1 + 7 1 7 7 We clam that + 1 + 7 1 7 7 ( 1+ For concseness, we let a ) ( 7 1 and b ) 7, observng that a b 7 Therefore, a b a m b m 7
Then, snce ab, and so whence a (1 b) 0 (mod b ) a m a(a ) (m 1)/ a (mod b ) a a b (mod b ), whch s a contradcton Hence, the sgn must be negatve Observe that we can rewrte the orgnal statement as m 7 (1 + 7) m (1 7) m, m ( m 7) ( 7) 0 Now consder each th teraton of the sum Suppose s odd Then the rght-hand sde becomes ( 7) ( 7) ( 7 + 7) ( 7) +1 Now suppose s even Then the rght-hand sde becomes ( 7) ( 7) Therefore, m 1 1 7 + 3 0 0 ( m 7 ± 5 ) ( 7 7) 7 m 1, m and we observe that m 1 m (mod 7) Now, 6 1 (mod 7), and t easly follows that the only solutons are then m 3, 5, or 13 (mod 4) We prove that only m 3, 5, and 13 can occur, and to prove unqueness t suffces to show that we cannot have two solutons of the orgnal equaton whch are congruent modulo 4 So let m, m 1 be two such solutons, and 7 l be the largest power of 7 dvdng m m 1 Then a m 1 a m a m 1 m a m (1/) m 1 m (1 + 7) m 1 m Now, and 1 m 1 [ (1 ) 6 ] m 1 m 6 1 (mod 7 l+1 ), (1 + 7) m 1 m 1 + (m 1 m) 7 (mod 7 l+1 ) 8
(frst rase to powers 7, 7,, 7 l, then (m m 1 )/7 Snce substtutng gves and a m 1 + m 7 m (mod 7), a m 1 a m + m 1 m 7 (mod 7 l+1 ), b m 1 b m m 1 m 7 (mod 7 l+1 ) But a m b m a m 1 b m 1, so (m 1 m) 7 0 (mod 7 l+1 ), but snce m 1 and m are ratonal ntegers, m 1 m (mod 7 l+1 ), contradctng the defnton of l Thus, m 3, 5, or 13, whch mples that n 5, 7, and 15, and solutons for x can easly be found The nterestng aspects of ths proof are the ponts of confluence between abstract algebra and number theory Whle the hypothess stated the unqueness of nteger solutons, the elements that were consdered belonged to an magnary quadratc feld In ths manner, many questons seemngly posed n number theory have proofs n ANT, such as Wle s famous proof of Fermat s Last Theorem n 1994 Ideas extendng prevous concepts are extremely useful to the progress of mathematcs, and should be embraced and studed extensvely as Algebrac Number Theory has been 4 Bblography [1]Stewart IN and DO Tall Algebrac Number Theory Chapman and Hall: London 1987 []Mollm, Rchard A Algebrac Number Theory Chapman and Hall: London 1999 [3]Pollard, Harry The Theory of Algebrac Numbers The Mathematcal Assocaton of Amerca: Baltmore 1950 [4]Ono, Takash An Introducton to Algebrac Number Theory Plenum Press: London 1990 5 Copyrght Ths work s lcensed under the Creatve Commons Attrbuton-NonCommercal 30 Unported Lcense To vew a copy of ths lcense, vst http://creatvecommonsorg/lcenses/by-nc/30/ 9