What Numbers Are of the Form a 2 +3b 2?

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1 What Numbers Are of the Form a +3b? Aryeh Zax 03/7/7 Fix an ambient UFD S. Forp S aprimeandn S \{0}, Denotebyv p (n) theexponent of p in the prime factorization of n, i.e. thelargestk such that p k n. Ourgoalistoshow the following result: Theorem. Fix n N. There is a solution of n = a +3b for a, b N if and only if for every prime p (mod3), v p (n) is even. Remark. This is in statement analogous to the more famous case n = a + b,wherethe condition is on p 3(mod4). Indeed,thebroadoutlineoftheproof(factoringinthe appropriate UFD extension of Z) isthesame. Lemma. The set of n of this form is closed under multiplication. Proof. (a +3b )(c +3d )=(ac 3bd) +3(ad + bc) Remark. This identity comes from the norm in Z[ p 3]. Lemma. Denote by! = +p 3 a primitive third root of unity. Then the Eisenstein integers, denoted by R = Z[!], are a Euclidean domain with the usual C norm. Remark. It s straightforward to check that Z[!] istwo-dimensionaloverz, withbasis{,!}. Proof of Lemma. Note that N(a + b!) =(a + b!)(a + b!) =a + b + ab(! +!) =a ab + b = a b b 0 This shows that the C-norm on Z[!] takesononlyintegervalues 0, with equality only when a = b =0. Itisalsoclearlymultiplicative;itremainstoshowthatitgivesrisetoa

2 Euclidean algorithm, i.e. given, R, 6= 0,wehavetofind, R with = + and N( ) <N( ). If then we re done, so suppose that -. Consider the R-ideal ( ), which forms a lattice inside C with fundamental region a rhombus. lies inside (or on the boundary of) one such rhombus. (Possibly more than one; in that case break ties arbitrarily.) That rhombus has sides of length ` = N( ) andinterioranglesof60 and 0. Draw circles of radius ` centered at the two vertices lying on the long diagonal, whose sectors are shown below. must be within ` from one of B or D, sincethesharedboundaryofthetwocirclesis {A, C}, which does not contain by hypothesis. Pick such that is the corresponding vertex (B or D), so = satisfies N( ) <N( ), as desired. Corollary. R is a UFD. Remark. We have that Z[ p 3] R, sinceforalla, b Z, a + b p 3=(a + b)+b + p 3 =(a + b)+b! Lemma 3. Let p>3 be an odd prime. Then there exists a solution x Z to p x +3 precisely when p (mod3). Proof (Quadratic Reciprocity). Observe that 3 3 p = =( ) (p )/ ( ) (p )/ = ( ) (p )/ p p = = p (mod 3) p p p Many thanks to my friend Rex Lei, whose suggestion significantly cleaned up my original proof.

3 Proof (Elementary). Since p is odd we can freely multiply and divide by ; let y Then x +3=(y ) +3=4y 4y +4=4(y y +) So x +3hasarootinF p i y y +does. =x. (=)) ifx +3hasaroot,thensodoesy y +;callitr. Itfollowsthaty + y +hasa root as well (namely r). Now r is a root of y 3, and r 6=, since ++=36= 0. Therefore r 6=,so r has order not and dividing 3, i.e. 3. Now 3 p bylagrange. ((=) If 3 p, then so does 6, and by the multiplicative group of a finite field being cyclic, there is an element of order 6, call it g. Now y 6 has6distinctroots(i.e.,g,g,,g 5 ). But we can write y 6 =(y 3 +)(y 3 ) = (y + y +)(y )(y y +)(y +) So two of those roots must be roots of y y +. This gives that x +3 also has a root. Corollary. If p Z is a prime, then p is a prime in R precisely when p (mod3). Proof. (=)) Weprovethecontrapositive.Ifp =3,then 3=( p 3)( p 3) witnesses that 3 is not prime. Otherwise, p (mod3)andwehavethatthereexistsx with p x +3=(x + p p 3)(x 3) but p - either factor on the right, so it is not prime. ((=) We show that all such p are irreducible. Suppose the contrary, and take p = a nontrivial factorization of p; sincen( )N( )=p and N( ) = =) =) is a unit we must have that N( ) =N( )=p. Write = x + y!, sothat N( ) = x y y = p Taking this equation modulo 3 yields (x y) +3y =4p (x y) p (mod 3) But all squares are either 0 or modulo 3, contradiction. Therefore all p (mod3)are irreducible in R. 3

4 Corollary. If p Z is a prime not equivalent to modulo 3, then p = a +3b for some a, b N. Proof. By the same line of argument as just used, we can write p = N(p) =N( )N( ) with N( ) =N( )=p. Writing = x + y! gives x y y = p () This is not exactly what we want (the LHS is not obviously of the form a +3b with a, b N), but it ll do. We proceed in two cases. Suppose x 6 y (mod ). Observe that and N(!) =N(!) =,so since we can freely permute x and y, itsu But then equation () can be written as x x + y! =!(y + x!) p = N(x + y!) =N(!)N(y + x!) =N(y + x!) y cestoconsiderthecaseofx odd and y even. y +3 = p and the LHS is of the form a +3b for a, b N, sowe redone. Suppose x y (mod ). Rewrite () as We want a solution to Supposing it exists, it satisfies (x y) +3y =4p c +3d = p 4p =(+3( ))(c +3d )=(c 3d) +3(c + d) and one can hope that c 3d =x y, c + d = y Solving this gives c = x + y,d= y x which are integers by hypothesis, so the result is shown. This hope is more plausible than it might seem; 4p has relatively few factors in Z[!], so there are only so many feasible ways to write it as a product of two of them. 4

5 We are finally ready to prove the main result: Theorem. Fix n N. There is a solution of n = a +3b for a, b N if and only if for every prime p (mod3), v p (n) is even. Proof. Write n = p e p e p e k k q f q f` `,wherethep i include 3 and all the primes that are mod 3 and the q i include all the primes that are modulo 3. (=)) Supposen = a +3b,andconsideranyprimep (mod3). Writen =(a + b p 3)(a b p 3). Since p is prime in R, p n implies that p a + b p 3orp a b p 3; since the two factors are conjugate and p is real, p therefore divides both of them. Evidently v p (a + b p 3) = v p (a b p 3), so v p (n) =v p (a + b p 3) is even. ((=) Suppose f i is even for all i, andwritem 0 = `Q and the set of such numbers is closed under multiplication, so m = i= q f i/ i. All the p i are of the desired form, some a, b. Now n = m(m 0 ) =(m 0 a) +3(m 0 b),andtheresultisshown. k Q j= p e i i = a +3b for 5