Irreducible Quartic Polynomials with Factorizations modulo p

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1 Irreducible Quartic Polynomials with Factorizations modulo Eric Driver, Phili A. Leonard, and Kenneth S. Williams 1. INTRODUCTION. In a discussion of irreducibility criteria in their fine algebra text, Dummit and Foote include the following remark [4,. 310]: Unfortunately, there are examles of olynomials even in Z[x] which are irreducible but whose reductions modulo every ideal are reducible... For examle, the olynomial x is irreducible in Z[x] but is reducible modulo every rime...and the olynomial x 4 72x is irreducible in Z[x] but is reducible modulo every integer. The matter is then ut on hold until Galois theory is treated five chaters later. The reason for the reducibility modulo of the cited olynomials is that they have the Klein 4-grou V as their Galois grous, so factorization modulo follows from the cycle structure of ermutations in this grou [4, Corollary 41,. 622]. But this and the related discussion of density [4,. 623 ff.] are much more advanced toics. We shall rovide an elementary treatment both of the examles cited and of families of related olynomials so that instructors might exlore this henomenon in a bit more detail earlier in the sequence of toics. In the course of doing so, we shall revisit several sources, including a recent Putnam Mathematical Cometition question and a few items from the ancient history of olynomial factorization. The question of reducibility modulo can be treated rather easily for the olynomials we consider. For certain families we discuss reducibility modulo every integer in some detail. 2. BIQUADRATIC POLYNOMIALS MODULO. Our attention in this section focuses on monic olynomials f of the form f (x) = x 4 + rx 2 + s, where r and s are integers. Although some authors use quartic and biquadratic as synonyms, we follow Kae and Warren [5] (among others) and reserve the latter term for olynomials of this secial tye. We ask which among them are irreducible over Z yet reducible modulo for each rime. We also investigate in section 3 the additional roerty of being reducible modulo n for every integer n larger than 1. It is convenient to begin by giving a necessary and sufficient condition for f (x) to be reducible in Z[x]. We rove: Theorem 1. Let r and s be integers. Then f (x) = x 4 + rx 2 + s is reducible in Z[x] if and only if there exist integers a, c, and e satisfying c + e a 2 r = 0, (1) a(e c) = 0, (2) 876 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 112

2 and ce s = 0, (3) in which case f (x) = (x 2 + ax + c)(x 2 ax + e). (4) Proof. Suose that f (x) is reducible in Z[x]. Then f (x) has either a linear factor x m in Z[x] or an irreducible quadratic factor x 2 + ax + c in Z[x]. In the former case, if m = 0then f (x) also has the second factor x + m, making it divisible by x 2 m 2 in Z[x], whereas if m = 0then f (x) is clearly divisible by x 2. In both cases f (x) has a quadratic factor x 2 + ax + c belonging to Z[x]. Thus x 4 + rx 2 + s = (x 2 + ax + c)(x 2 + tx + e) for integers t and e. Equating coefficients of x 3, we conclude that t = a and (4) follows. Equating coefficients of x 2, x, and 1, we obtain (1), (2), and (3), resectively. Conversely, suose that (1), (2), and (3) hold. Then (x 2 + ax + c)(x 2 ax + e) = x 4 + (c + e a 2 )x 2 + a(e c)x + ce = x 4 + rx 2 + s, whence f (x) is reducible in Z[x]. We now continue to develo the reducibility criteria for f (x). In what follows, we often write n = to indicate that an integer n is a erfect square. Since r 2 4s is the discriminant of the quadratic f (x 1/2 ), it is reasonable to susect that whether r 2 4s is a erfect square or not will have an effect on if and how f (x) reduces. Indeed, the next two corollaries to Theorem 1 show how the factorization of f (x) deends on the quantity r 2 4s. In articular, when r 2 4s =,then f (x) factors into a roduct of two quadratics with no linear terms; and when r 2 4s = and f (x) is reducible, it must factor into two quadratics with linear terms. Corollary 1. Let r and s be integers such that r 2 4s is a erfect square, say r 2 4s = t 2 (t Z).Then f(x) = x 4 + rx 2 + s is reducible in Z[x], and f (x) = ( x (r t))( x (r + t)). Corollary 2. Let r and s be integers such that r 2 4s is not a erfect square. Then f (x) = x 4 + rx 2 + s is reducible in Z[x] if and only if there exists an integer c such that in which case c 2 = s, 2c r =, f (x) = (x 2 + ax + c)(x 2 ax + c), where a is an integer such that a 2 = 2c r. December 2005] IRREDUCIBLE QUARTIC POLYNOMIALS 877

3 As Corollaries 1 and 2 are simle consequences of Theorem 1, we leave their roofs to the reader. We remark that, if r and s are integers such that r 2 4s = and if x 4 + rx 2 + s is reducible in Z[x], then the integer c of Corollary 2 satisfies so that (2c r)( 2c r) = r 2 4c 2 = r 2 4s =, 2c r = 0, 2c r =. Since ±c are the only integers for which c 2 = s, this shows that c is unique. The olynomials cited in the introduction rovide good illustrations for this result. In the case of x 4 + 1wehaver = 0, s = 1, and r 2 4s = 4 =. Because there does not exist an integer c satisfying c 2 = 1and2c =, we conclude on the basis of Corollary 2 that x is irreducible in Z[x]. With x 4 72x we reach the same conclusion: here r = 72, s = 4, and r 2 4s = 5168 =, and there does not exist an integer c such that c 2 = 4and2c + 72 =. A nice examle of a family of biquadratics for which reducibility in Z[x] is easily decided was given in a recent Putnam Cometition [8, Problem A3], where the olynomials P m (x) = x 4 (2m + 4)x 2 + (m 2) 2 (m Z) were introduced. Here r = (2m + 4) and s = (m 2) 2, which means that r 2 4s = (2m + 4) 2 4(m 2) 2 = 32m is a erfect square if and only if 2m is a erfect square. If 2m is a square, then m = 2u 2 for some integer u and, by Corollary 1, we have P m (x) = (x 2 (2u 2 4u + 2))(x 2 (2u 2 + 4u + 2)). If 2m is not a erfect square then, again by Corollary 2, P m (x) is reducible if and only if ±2(m 2) + 2m + 4 =, which haens if and only if m = t 2 for some t in Z,in which case P m (x) = (x 2 + 2tx + t 2 2)(x 2 2tx + t 2 2). Thus P m (x) is irreducible in Z[x] excet when m or 2m is a erfect square. So far we have considered the reducibility criteria for f (x) = x 4 + rx 2 + s over Z. We now turn our attention to the other half of the roblem, namely, to reducibility criteria for f (x) over Z/ k Z,where is a rime and k is a ositive integer. Our next theorem is the direct analog of Theorem 1, and its roof is similar enough to the roof of that result to be left to the reader. Theorem 2. Let be a rime, and let k be a ositive integer. Then a olynomial f (x) = x 4 + rx 2 + sinz[x] is reducible modulo k if and only if there exist integers a, c, and e satisfying c + e a 2 r 0 (mod k ), (5) a(e c) 0 (mod k ), (6) 878 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 112

4 and ce s 0 (mod k ), (7) in which case f (x) (x 2 + ax + c)(x 2 ax + e)(mod k ). (8) We now give the comlete story of the factorization of f (x) modulo an odd rime. Two secial cases are easy to handle. If s 0 (mod ), then (5), (6), and (7) are solvable with k = 1, a = c = 0, and e = r, imlying that x 4 + rx 2 + s x 2 (x 2 + r)(mod ) is reducible. If r 2 4s 0 (mod ) then (5), (6), and (7) are solvable with k = 1, a = 0, and c = e r/2 (mod ),so x 4 + rx 2 + s x 4 + rx 2 + r 2 /4 (x 2 + r/2) 2 (mod ) is reducible. (Here 1/2 signifies the inverse of 2 modulo.) In describing the rest of the criteria, we use roerties of the Legendre symbol (k/), which is defined for an integer k and an odd rime by ( ) k = 1if k and x 2 k (mod ) is solvable, 0if k, 1 if k and x 2 k (mod ) is insolvable. It is well known that for all integers k and l, andthat ( ) ( )( ) kl k l = ( ) ( ) k 2 l l = if k. The following result is due to Carlitz [1] and can be deduced in an elementary way from Theorem 2: Theorem 3. If is an odd rime and if r and s are integers such that s 0 (mod ) and r 2 4s 0 (mod ), then the following statements hold: (i) f (x) = x 4 + rx 2 + s is the roduct of two distinct monic linear olynomials and an irreducible monic quadratic olynomial modulo if and only if ( ) ( ) s r 2 4s = 1, = 1. (9) (ii) f (x) = x 4 + rx 2 + s is the roduct of four distinct monic linear olynomials modulo if and only if December 2005] IRREDUCIBLE QUARTIC POLYNOMIALS 879

5 ( ) s = 1, ( ) r 2 4s = 1, ( ) r 2t = 1, (10) where t is an integer such that s t 2 (mod ). (iii) f (x) = x 4 + rx 2 + s is the roduct of two distinct monic irreducible quadratic olynomials modulo if and only if ( ) s = 1, ( ) r 2 4s = 1, ( ) r 2t = 1, (11) where t is an integer such that s t 2 (mod ),or ( ) s = 1, ( ) r 2 4s = 1. (12) (iv) f (x) = x 4 + rx 2 + s is irreducible modulo if and only if ( ) s = 1, ( ) r 2 4s = 1. (13) As a bridge from Theorem 3 to the next result, we consider the irreducibility of f (x) = x 4 10x modulo,where is again an odd rime. By Theorem 3 (iv) this olynomial is irreducible modulo if and only if that is, if and only if ( ) 17 = ( ) 2 = ( ) 32 = 1, ( ) 17 = 1. In articular, it is irreducible modulo 61. This accounts for the relacement of x 4 10x [3,. 303] with x 4 72x in the second edition [4,. 310], and oints to the condition contained in the next theorem. It is worth noting that, u until this oint, all of our arguments have involved only elementary results from number theory. However, the roof of the next result relies on Dirichlet s famous theorem on rimes in arithmetic rogressions, which states that there are infinitely many rimes with a (mod m) for anya relatively rime to m. Theorem 4. Let r and s be integers such that r 2 4s is not a erfect square. Then the olynomial f (x) = x 4 + rx 2 + s is reducible modulo for every rime if and only if s = t 2 for some integer t. Proof. Suose that s = t 2 for an integer t. Consider an odd rime.ifs 0 (mod ) or r 2 4s 0 (mod ) then, by the remarks following Theorem 2, f (x) is reducible modulo. Ifs 0 (mod ) and r 2 4s 0 (mod ), then(s/) = 1, so f (x) is reducible modulo (Theorem 3(ii),(iii)). Thus f (x) is reducible modulo every odd rime. For = 2, there are only the four biquadratic olynomials x 4, x 4 + 1, x c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 112

6 x 2,andx 4 + x to consider modulo 2, and it is easy to check that each of these is reducible modulo 2. Conversely, suose that f (x) is reducible modulo for every rime. Suose that s is not a erfect square. Then, as both r 2 4s and s are nonsquares, by the law of quadratic recirocity and Dirichlet s theorem on rimes in arithmetic rogressions, there exist infinitely many rimes such that ( ) r 2 4s = ( ) s = 1. In view of Theorem 3(iv), f (x) is irreducible modulo each of these rimes, a contradiction. Hence s is a erfect square. We are now in a osition to state the main result of this section, which gives necessary and sufficient conditions for a olynomial to be irreducible over Z but reducible modulo for every rime. This result is easily obtained by combining Corollaries 1 and 2 with Theorem 4. Theorem 5. Let r and s be integers. Then the olynomial f (x) = x 4 + rx 2 + sis irreducible in Z[x] but is reducible modulo for every rime if and only if the following are true: r 2 4s =, s =, 2 s r =, 2 s r =. We return again to the olynomials cited in the introduction. First, for the olynomial f (x) = x 4 + 1, we have r = 0, s = 1, r 2 4s = 4 =, s = 1, 2 s r = 2 =, 2 s r = 2 =. Aealing to Theorem 5, we see that f (x) is irreducible in Z[x] but is reducible modulo for every rime. In the case of the olynomial f (x) = x 4 72x 2 + 4, r = 72, s = 4, r 2 4s = 5168 =, s = 2, 2 s r = 76 =, 2 s r = 68 =. Thus, by Theorem 5, f (x) is irreducible in Z[x] but is reducible modulo for every rime. As another examle consider f (x) = x 4 10x Here we have r = 10, s = 1, r 2 4s = 96 =, s = 1, 2 s r = 12 =, 2 s r = 8 =. According to Theorem 5, f (x) is irreducible in Z[x] but is reducible modulo for every rime. Next consider the family of olynomials F a defined by F a (x) = x 4 + 2(1 a)x 2 + (1 + a) 2 (a Z). December 2005] IRREDUCIBLE QUARTIC POLYNOMIALS 881

7 Here r = 2(1 a), s = (1 + a) 2, r 2 4s = 16a, s = 1 + a, 2 s r = 4a, 2 s r = 4. Theorem 5 imlies that F a (x) is irreducible in Z[x] but is reducible modulo for every rime if and only if neither a nor a is a square. This is a slight generalization of Lee s theorem [6]. As yet another examle, we revisit the Putnam roblem, where P m (x) = x 4 (2m + 4)x 2 + (m 2) 2 (m Z). Suose that m = and 2m =. Then from what we showed earlier, P m (x) is irreducible in Z[x]. By Theorem 4 P m (x) is reducible modulo for every rime. We now give an examle of a one-arameter family of biquadratic olynomials that are irreducible in Z[x] and reducible modulo for every rime for every value of the arameter. Let L k (x) = x 4 (4k + 1)x (k Z). In this situation r = 4k 1, s = 1, r 2 4s = 16k 2 + 8k 3. As a square is congruent to 0, 1, or 4 modulo 8, we see that r 2 4s, 2 s r, and 2 s r are not erfect squares, since r 2 4s 5 (mod 8), ±2 s r 3 (mod 4). Hence, by Theorem 5, the olynomial L k (x) is irreducible in Z[x] but reducible modulo for every rime indeendently of k. We close this section by noting that, when a biquadratic olynomial x 4 + rx 2 + s in Z[x] is irreducible, its Galois grou is the Klein 4-grou V if and only if s is a erfect square. This follows, for examle, from [5, Theorem 1(iii),. 134]. Thus the biquadratic olynomials x 4 + 1, x 4 10x 2 + 1, x 4 72x 2 + 4, P m (x) (m, 2m = ), F a (x) (±a = ), L k (x) (k Z) all have the Klein 4-grou V as their Galois grous. On the other hand, the biquadratic olynomial x 4 10x has the dihedral grou D 4 of order 8 as its Galois grou. It is the fact that the latter grou contains elements ermuting the roots of x 4 10x in a 4-cycle, together with the Tchebotarov density theorem, that accounts for the existence of (infinitely many) rimes for which the olynomial is irreducible modulo [4,. 623]. 3. BIQUADRATIC POLYNOMIALS MODULO n. We have seen that both x and x 4 72x are irreducible in Z[x] and reducible modulo for every rime. These two differ, however, when we consider their reducibility modulo an arbitrary ositive integer n. If we suose that x is reducible modulo 4, then Theorem 2 establishes the existence of integers a, c, ande such that c and e are odd and c e (mod 4). But 882 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 112

8 then a 2 c + e 2c 2 (mod 4), which is imossible. Thus x is not reducible modulo 4. On the other hand, we claim that f (x) = x 4 72x is reducible modulo n for every ositive integer n larger than 1. We recall two facts from elementary number theory: if a is an integer such that a 1 (mod 8), then the congruence x 2 a (mod 2 k ) is solvable for every ositive integer k;if is an odd rime and b is an integer such that (b/) = 1, then the congruence x 2 b (mod k ) is solvable for every ositive integer k. By the Chinese remainder theorem it suffices to rove that f (x) is reducible modulo k for each ositive integer k and each rime. We consider a number of cases. Case 1: = 2. Since 17 1 (mod 8), there exists an integer A with A 2 17 (mod 2 k ), whence f (x) (x 2 + 2Ax 2)(x 2 2Ax 2)(mod 2 k ). Case 2: = 17. Because (19/17) = (2/17) = 1, there exists an integer B such that B 2 19 (mod 17 k ),so f (x) (x 2 + 2Bx + 2)(x 2 2Bx + 2)(mod 17 k ). Case 3: = 19. As (17/19) = ( 2/19) = 1, there exists an integer C satisfying C 2 17 (mod 19 k ). This gives rise to the factorization f (x) (x 2 + 2Cx 2)(x 2 2Cx 2)(mod 19 k ). Case 4: = 2, 17, or 19 and (17/) = 1. In this case there exists an integer D such that D 2 17 (mod k ) and f (x) (x 2 + 2Dx 2)(x 2 2Dx 2)(mod k ). Case 5: = 2, 17, or 19 and (19/) = 1. Here there exists an integer E with E 2 19 (mod k ), imlying that f (x) (x 2 + 2Ex + 2)(x 2 2Ex + 2)(mod k ). Case 6: = 2, 17, or 19 and (17/ ) = (19/ ) = 1. We have (323/ ) = (17/)(19/) = 1, so there exists an integer F for which F (mod k ).In this instance f (x) (x F)(x F)(mod k ). It is now time to see what distinguishes an irreducible olynomial like x 4 + 1, which is reducible modulo for every rime but not modulo n for every integer n greater than 1, from the irreducible olynomial x 4 72x 2 + 4, which is reducible modulo n for n = 2, 3, 4,... We consider a biquadratic olynomial f (x) = x 4 + rx 2 + s in Z[x] satisfying the conditions of Theorem 5, ensuring that f (x) is irreducible in Z[x] and reducible modulo for every rime. We describe the recise conditions under which f (x) is also reducible modulo n for every integer n bigger than 1. By the Chinese remainder theorem, it suffices to determine the conditions under which f (x) is reducible modulo every rime ower k. December 2005] IRREDUCIBLE QUARTIC POLYNOMIALS 883

9 Theorem 6. Let f (x) = x 4 + rx 2 + s be a olynomial in Z[x] such that the following hold: r 2 4s =, s =, 2 s r =, 2 s r =. Write s = t 2 for an integer t, and let be a rime. Then f (x) is reducible modulo k for every ositive integer k if and only if r 2 4t 2 is a square modulo k for every ositive integer k or r + 2t is a square modulo k for every ositive integer k or r 2t is a square modulo k for every ositive integer k. The roof of Theorem 6, although straightforward, is somewhat technical, so we defer it to the aendix at the end of the article. In order to aly Theorem 6, one requires conditions under which a nonzero integer a is a square modulo k for a given rime and all ositive integers k. From results in elementary number theory (see, for examle, [7, ] for a concise treatment of what is needed), we have the following: Fact 1. Let a be a nonzero integer, and write a = 2 m a 0,wherea 0 is an odd integer and m is a nonnegative integer. Then x 2 a (mod 2 k ) has solutions for all ositive integers k if and only if m is even and a 0 1 (mod 8). Fact 2. Let a be a nonzero integer, let be an odd rime, and write a = m a 0, where a 0 is an integer not divisible by and m is a nonnegative integer. Then x 2 a (mod k ) has solutions for all ositive integers k if and only if m is even and a 0 is a quadratic residue modulo. We need these results in order to exhibit a class of irreducible biquadratic olynomials x 4 + rx 2 + s that are reducible modulo n for every ositive integer n greater than 1. Theorem 7. If q 1 and q 2 are distinct odd rimes such that ( ) q2 q 1 1(mod8), = 1, and if f (x) = x 4 2(q 1 + q 2 )x 2 + (q 1 q 2 ) 2,then f(x) is irreducible in Z[x] but reducible modulo n for n = 2, 3, 4,... Proof. Here r = 2(q 1 + q 2 ), s = (q 1 q 2 ) 2,andt = q 1 q 2. Thus r 2 4t 2 = 16q 1 q 2 =, r + 2t = 4q 1 =, r 2t = 4q 2 =. Consider an arbitrary ositive integer k. Asq 1 1 (mod 8), Fact 1 ensures that the integer r + 2t is a square modulo 2 k. Because (q 2 /q 1 ) = 1, by Fact 2 r 2t is a square modulo q1 k.nowq 1 1 (mod 8) and (q 2 /q 1 ) = 1, so by the law of quadratic recirocity (q 1 /q 2 ) = 1. Fact 2 tells us that r + 2t is a square modulo q2 k.if is a rime different from 2, q 1,orq 2 such that (q 1 /) = 1, then r + 2t is a square modulo k. Similarly, if (q 2 /) = 1, r 2t is a square modulo k, while if (q 1 /) = (q 2 /) = 1, then (q 1 q 2 /) = 1andr 2 4t 2 is a square modulo k. Hence, f (x) is irreducible in Z[x] and reducible modulo k for every rime and every ositive integer k (Theorem 6), and by the Chinese remainder theorem f (x) is reducible modulo n for each ositive integer n greater than 1. q c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 112

10 The olynomial x 4 72x 2 + 4arisesfromTheorem7bytakingq 1 = 17 and q 2 = 19. We note that, since there are infinitely many rimes q 2 with q 2 1 (mod 2q 1 ) for any rime q 1 such that q 1 1 (mod 8), the class of olynomials in Theorem 7 is infinite. 4. STICKELBERGER S PARITY THEOREM. The ancient history of olynomial factorization contains a result that exlains the reducibility modulo of olynomials such as x 4 + 1andx 4 72x Originally due to Stickelberger [10], it has been rediscovered and studied in more recent times (see [2] and[11]). We state Stickelberger s theorem in a form convenient for our uroses and encourage the reader to consult Swan s aer [11] for a roof and some interesting alications. Theorem 8. Let f (x) be a monic olynomial of degree n in Z[x], and let be an odd rime not dividing the discriminant D( f ) of f. Suose that f (x) f 1 (x) f 2 (x) f r (x)(mod ), where the f j are irreducible olynomials modulo. Then n r (mod 2) if and only if (D( f )/) = 1. As is immediately clear, this result indicates that a olynomial of even degree with (nonzero) square discriminant is always factorable modulo for all odd rimes not dividing its discriminant. 5. A FAMILY FOR THE ALTERNATING GROUP. In discussing quartics irreducible over the integers but reducible modulo every rime, we have treated biquadratics in some detail. These olynomials have the Klein 4-grou as their Galois grous. In this section we introduce a family of olynomials having the alternating grou A 4 as their Galois grous, and we invite the reader to exlore the same henomena for these olynomials. The ancient history yields a starting oint. Seidelmann [9] determined the quartics with rational coefficients having the alternating grou A 4 as their Galois grous, namely, [e 3 ( f 2 + 3g 2 )(3e + 2 f )]x 4 6ex 2 e 2 4 f 2 12g 2 8x 3 e 3 ( f 2 + 3g 2 )(3e + 2 f ), with e, f,andg rational such that e 3 ( f 2 + 3g 2 )(3e + 2 f ) is not zero. Setting f = e 1 and g = 1, we obtain the olynomials 2 2 f e (x) = x 4 6ex 2 8x + (9e e + 12) (e Z). Each of these can be shown to be irreducible by establishing that (i) f e (x) has no linear factors in Z[x], and (ii) f e (x) has no quadratic factors in Z[x]. We give the details only for (ii). If f e (x) has a quadratic factor, then it admits a factorization of the form x 4 6ex 2 8x + (9e e + 12) = (x 2 + Ax + B)(x 2 Ax + C) December 2005] IRREDUCIBLE QUARTIC POLYNOMIALS 885

11 for integers A, B, andc. Comaring coefficients leads to the three equations: Examining the owers of 2 in (15), we see that A 2 + B + C = 6e, (14) A(C B) = 8, (15) BC = 9e e (16) A = ε2 r, B C = ε2 3 r, where ε = 1or1andr belongs to {0, 1, 2, 3}. From (14), we deduce that B + C = 2 2r 6e. Solving the two linear equations for B and C shows that r is 1 or 2 and that B = 2 2r 1 3e + ε2 2 r, C = 2 2r 1 3e ε2 2 r. Inserting these values of B and C into (16), we arrive at (2 2r 1 3e) r = 9e e However, for r = 1 or 2 this equation yields only nonintegral values for e, a contradiction. Recall now (see, for examle, [4,. 595]) that the discriminant D of a quartic olynomial x 4 + Px 2 + Qx + R is given by D = 16P 4 R 4P 3 Q 2 128P 2 R PQ 2 R 27Q R 3. (17) We infer that the discriminant D e of f e (x) is D e = (e 2 + e + 1) 2. Since D e is a erfect square, Theorem 8 imlies that, for any odd rime not dividing D e, f e (x) has an even number of irreducible factors modulo andsoisreducible modulo.for = 2wehave for = 3 and for ( = 2, 3) dividing e 2 + e + 1 f e (x) (x + e) 4 (mod 2), f e (x) x(x + 1) 3 (mod 3), f e (x) (x + 3(e + 1))(x (e + 1)) 3 (mod ). Thus f e (x) is reducible modulo for every rime. For the sake of comleteness, we aeal to [5] to show that the Galois grou of f e (x) is A 4 for every integer e. As the discriminant of f e (x) is a erfect square, we know that its Galois grou must be a subgrou of the alternating grou A 4 [4, Proosition 34,. 592]. We recall that if the quartic olynomial x 4 + Px 2 + Qx + R has 886 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 112

12 roots α 1, α 2, α 3,andα 4 in C, then the cubic olynomial with the roots α 1 α 2 + α 3 α 4, α 1 α 3 + α 2 α 4,andα 1 α 4 + α 2 α 3 is the resolvent cubic of x 4 + Px 2 + Qx + R. Itis given by x 3 Px 2 4Rx + (4PR Q 2 ) (see, for examle, [5]). In order to rove that the Galois grou of f e (x) is actually A 4 we must show that the resolvent cubic of f e (x) has no roots in Z [5, Theorem 1(ii),. 134]. The resolvent cubic in question is g e (x) = x 3 + 6ex 2 4(9e e + 12)x 8(3e(9e e + 12) + 8). Suose that m is an integral root of g e (x). Clearly m must be even, say m = 2n, and (after dividing by 8) we have It follows that n 3 + 3en 2 (9e e + 12)n (3e(9e e + 12) + 8) = 0. (n + 3e)n 2 (n + 3e)(9e e + 12) 8 = 0, (18) from which it is immediate that n + 3e divides 8, so n + 3e = ε2 w with ε in { 1, 1} and w in {0, 1, 2, 3}. Substituting n = ε2 w 3e into (18) yields (ε2 w 6e)2 2w 12(e + 1)ε2 w 8 = 0. The three ossibilities w = 0, 2, or 3 are easily ruled out by divisibility considerations. It remains to consider the situation for w = 1. In this case, after substituting and then dividing by 8, we obtain whichinturnyields ε 3e 3eε 3ε 1 = 0, 2ε 1 = 3e(ε + 1). This equation shows that ε = 1. Thus ε = 1, which leads to e = 1/2, contradicting the fact that e is an integer. Therefore the olynomial g e (x) has no integer roots, whence the Galois grou of f e (x) is A 4. We leave the reader with the question: Are there integers e for which f e (x) is reducible modulo n for every integer n greater than 1? 6. APPENDIX: PROOF OF THEOREM 6. ( ) Suose first that u 2 r 2 4t 2 (mod k ) has a solution for every ositive integer k, where is an odd rime. Let w denote the inverse of 2 modulo k.then f (x) is reducible modulo k,for f (x) (x 2 + w(r + u))(x 2 + w(r u)) (mod k ). Suose next that u 2 r 2 4t 2 (mod 2 h ) has a solution for every ositive integer h. Letk be a ositive integer, and let v be a solution of v 2 r 2 4t 2 (mod 2 k+2 ), so v r (mod 2). Then s = 1 (v r) is an integer. Clearly v = r + 2s. Alsos(r + s) is 2 congruent to t 2 modulo 2 k. Hence f (x) is reducible modulo 2 k : namely, f (x) x 4 + rx 2 s(r + s) (x 2 s)(x 2 + r + s)(mod 2 k ). Now assume that u 2 r + 2t (mod k ) has a solution for each ositive integer k, where is an arbitrary rime. Then f (x) is reducible modulo k with the factorization f (x) (x 2 + ux + t)(x 2 ux + t)(mod k ). December 2005] IRREDUCIBLE QUARTIC POLYNOMIALS 887

13 Finally, when u 2 r 2t (mod k ) has a solution for each rime and each ositive integer k,then f (x) has the factorization f (x) (x 2 + ux t)(x 2 ux t)(mod k ), revealing that it is reducible modulo k. ( ) Suose that f (x) is reducible modulo k for every ositive integer k.fixsuch a k. We first consider the case when is an odd rime not dividing r 2 4t 2.If ( ) r 2 4t 2 = 1, then r 2 4t 2 is a square modulo k.if ( ) r 2 4t 2 = 1, then so or ( )( ) r + 2t r 2t = 1, ( ) r + 2t = 1 ( ) r 2t = 1. Accordingly, either r + 2t or r 2t is a square modulo k. Assume next that is an odd rime dividing r 2 4t 2. Because r 2 4t 2 =, we deduce that t = 0. Exress t as t = l t 0,wherel is a nonnegative integer and t 0 is an integer not divisible by, andletk be a ositive integer. Since f (x) is reducible modulo 2k+4l, Theorem 2 declares the existence of integers a, c,ande such that c + e a 2 r 0 (mod 2k+4l ), (19) a(c e) 0 (mod 2k+4l ), (20) ce s 0 (mod 2k+4l ). (21) By adding 2k+4l to a, if necessary, we may suose that a = 0. Similarly adding 2k+4l to c, if necessary, we may suose that c = e. Let k 1 and k 2 be the exact owers of dividing a and c e, resectively. Then, by (20), k 1 + k 2 2k + 4l. Suose first that k 1 k + 2l. Thena 2 0(mod 2k+4l ). From (19) we infer that so (21) leads to c + e r (mod 2k+4l ), r 2 4t 2 = r 2 4s (c + e) 2 4ce (c e) 2 (mod 2k+4l ). 888 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 112

14 Thus u 2 r 2 4t 2 (mod k ) is solvable for each k in N. Now assume that k 1 < k + 2l. Thenk 2 2k + 4l k 1 > k + 2l. Set Referring to (19), we conclude that e c = d 0 k 2, d 0. 2c + d 0 k 2 a 2 r = (c + e a 2 r) (e c d 0 k 2 ) 0 (mod 2k+4l ). As a result and, since k 2 k + 2l + 1, Invoking (21), we obtain which imlies that for ε = 1or 1. Finally, r + 2c a 2 d 0 k 2 (mod 2k+4l ) r + 2c a 2 (mod k+2l+1 ). 2l t 2 0 = t 2 = s ce c(c + d 0 k 2 ) c 2 (mod k+2l+1 ), t = l t 0 εc(mod k+1 ) r + 2εt a 2 (mod k ). The case = 2 can be treated in a similar manner. We leave the details to the reader. REFERENCES 1. L. Carlitz, Note on a quartic congruence, this MONTHLY 63 (1956) K. Dalen, On a theorem of Stickelberger, Math. Scand. 3 (1955) D. S. Dummit and R. M. Foote, Abstract Algebra, Prentice-Hall, Englewood Cliffs, N.J., , Abstract Algebra, 2nd ed., Prentice-Hall, Uer Saddle River, N.J., L.-C. Kae and B. Warren, An elementary test for the Galois grou of a quartic olynomial, this MONTHLY 96 (1989) M. A. Lee, Some irreducible olynomials which are reducible mod for all, thismonthly 76 (1969) W. J. LeVeque, Toics in Number Theory, vol. 1, Addison-Wesley, Reading, MA, nd Annual William Lowell Putnam Mathematical Cometition, Math. Magazine 75 (2002) F. Seidelmann, Die Gesamtheit der kubischen und biquadratischen Gleichungen mit Affekt bei beliebegem Rationalitätsbereich, Math. Annalen 78 (1918) L. Stickelberger, Über eine neue Eigenschaft der Diskriminanten algebraischer Zahlkörer, in Verhandlungen des I Internationalen Mathematiker-Kongresses,Zürich, August 1897, F. Rudio, ed., B. G. Teubner, Leizig, 1898, R. G. Swan, Factorization of olynomials over finite fields, Pacific J. Math. 12 (1962) December 2005] IRREDUCIBLE QUARTIC POLYNOMIALS 889

15 ERIC DRIVER received his M.S.E and Ph.D. degrees in electrical engineering from Arizona State University in 1993 and 1996, resectively. He has worked for Lockheed Martin in Goodyear, Arizona, for the last eight years, designing advanced radar systems. He has always loved mathematics, esecially number theory, and is resently ursuing a Ph.D. in mathematics at Arizona State University. His research interests include algebraic number theory, class field theory, and algebraic toology. 700 E. Mesquite Circle N128, Teme, AZ eric.driver@lmco.com PHILIP A. LEONARD recently retired from Arizona State University, where he served for thirty-six years. His interests have been in number theory and asects of combinatorics, as well as in working with students rearing to teach mathematics. He received the 2002 MAA Southwestern Section Award for Distinguished College or University Teaching of Mathematics. Deartment of Mathematics, Arizona State University, Teme, AZ hili.leonard@asu.edu KENNETH S. WILLIAMS finished his Ph.D. in mathematics at the University of Toronto in After a year as a lecturer at the University of Manchester, England, he joined the Deartment of Mathematics at Carleton University in Ottawa. He became a full rofessor in 1975 and served as chair of the deartment from 1980 to 1984 and again from 1997 to In 1998, when the deartment became the School of Mathematics and Statistics, he became its first director, serving until In 2002 he retired as rofessor emeritus and distinguished research rofessor. He continues to suervise the theses of graduate students in number theory. His hobbies include birding, hiking and gardening. School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada K1S 5B6. williams@math.carleton.ca I even know of a mathematician who slet with his wife only on rimenumbered days..., Graham said. The Man Who Loved Only Numbers by Paul Hoffman A mathematician was obsessed with things rime. He thought about them almost all of the time. Said to his dear wife, It truly seems right That we should only make love on a rime-numbered night. His wife thought for a bit ( cause she was no dummy), At the month s start this does seem quite yummy, For there s two, three, five, seven A three-night hiatus and then there s eleven. But of the month s end I start to be wary Near the twenty-third day of the month February. For the next rime day after will be March the first Such sexual continence might cause me to burst! He shook his head sadly, As it s commonly reckoned, The next rime day would be found on the second. Submitted by John Drost, Marshall University 890 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 112