Polynomial PSet Solutions

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1 Polynomal PSet Solutons Note: Problems A, A2, B2, B8, C2, D2, E3, and E6 were done n class. (A) Values and Roots. Rearrange to get (x + )P (x) x = 0 for x = 0,,..., n. Snce ths equaton has roots x = 0,,..., n, and the left-hand sde has degree n+, (x + )P (x) x = kx(x ) (x n) for some nteger k. Then P (x) = kx(x ) (x n) + x. x + The numerator must be dvsble by x + so must have x = as a zero. Pluggng n, we get ( ) n+ (n + )!k = 0, or k = ( )n+. Pluggng n x = n + gves (n+)! {, n odd P (n + ) = n, n even n+2 2. Ths tme t s easer to guess the soluton. The polynomal Q(x) = n =0 ( ) x (r ) has degree n and by the Bnomal Theorem, satsfes the gven condtons. Snce P (x) = Q(x) for n + values of x, actually P, Q are the same polynomal, and P (n + ) = Q(n + ) = ( n+ ( ) n + )(r ) (r ) n+ = r n+ (r ) n+. =0 3. Lettng ray OQ be the postve real axs, Q represent the nth roots of unty ω n the complex plane. Hence P Q equals 2 ω. The roots of x n = 0 are just the nth roots of unty, so x n = n =0 (x ω ). Pluggng n x = 2 gves n k= P Q = 2 n. 4. The gven condton says f(x) 2 = x(x ) (x n)q(x) + x 2 + () for some polynomal f(x) of degree at most n. Pluggng x = 0,,..., n nto () gves f(x) = ± x 2 +, when x = 0,,..., n. (2)

2 The followng s key: Gven n + ponts (x 0, y 0 ),..., (x n, y n ) wth dstnct x- coordnates, there exsts exactly one polynomal f of degree at most n so that f(x ) = y for = 0,,..., n. Applyng ths to (2) we get 2 n+ possbltes for f(x) snce we have 2 choces of sgn for each of x = 0,,..., n. If f(x) s a soluton to (2) then so s f(x); we get 2 n possbltes for f(x) 2. Solve () to get 2 n possbltes for Q(x): Q(x) = f(x)2 x 2 x(x ) (x n) Each such polynomal s a vald soluton because f(x) 2 x 2 s zero at x = 0,,..., n and hence s dvsble by x(x ) (x n). 5. Clearng denomnators, [ 5 a x = j, j 5 (x + j) ] (x + )(x + 2)(x + 3)(x + 4)(x + 5) = 0 for x =, 4, 9, 6, 25. Let f(x) denote the LHS. Snce f(x) = 0 has the roots x =, 4, 9, 6, 25, we conclude that (x )(x 4) (x 25) dvdes f(x). Snce f(x) has degree at most 5, f(x) = k(x )(x 4)(x 9)(x 6)(x 25) for some constant k. However, equatng f(0) = [ (x + )(x + 2)(x + 3)(x + 4)(x + 5)] x=0 = 5! and f(0) = k 5! 2 gves k = 5!. Thus Then 5 = f(x) = (x )(x 4)(x 9)(x 6)(x 25). 5! a = f(6 2 ) x(x + )(x + 2)(x + 3)(x + 4)(x + 5) x=6 2 = Suppose p(x) r x < for x = 0,,..., n. Let y = p() for 0 n. By the Lagrange Interpolaton Formula, p(x) = [ n y =0 2 j ] x j. j

3 When r y r +, p(n + ) s maxmzed when we take y = r when the product s negatve (.e. when n s odd) and y when the product s postve (.e. when n s even): n p(n + ) < (r + ( ) n ) (n + ) j j =0 j n = (r + ( ) n )( ) n (n + )!!(n + )! =0 n [( ) ( )] n + n + = r ( ) n + =0 = ((r ) n+ r n+ ) + (2 n+ ) by the Bnomal Theorem = r n+ + 2 n+ (r ) n+ r n+ An alternate soluton s to use Lagrange Interpolaton wth n + 2 ponts and show that the coeffcent of x n+ cannot be Gven two ponts (x, y ) and (x 2, y 2 ) the equaton y y y 2 y = x x x 2 x = (y y )(x 2 x ) = (x x )(y 2 y ) represents the lne passng through these two ponts (t s a lnear equaton satsfed by the coordnares of the two ponts). It follows that three ponts (x, y ), (x 2, y 2 ) and (x 3, y 3 ) le on the same lne f and only f the condton (y 3 y )(x 2 x ) = (x 3 x )(y 2 y ) (3) holds. Now suppose that (x, y ), (x 2, y 2 ) and (x 3, y 3 ) represent the (changng) coordnates of the three ducks as they waddle along ther paths. Each coordnate s a lnear functon of tme t, so (3) s an equaton n t of degree at most 2 (.e., s a quadratc). If such an equaton has more than 2 solutons then t must reduce to an dentty and thus hold true for all values of t. That s, f the ducks are n a row at more than two tmes, then they are always n a row. 8. Note that g(x) s one of the 6 nteger dvsors of 2008 for each of the 8 nteger roots. There must be at least 6 roots of f(x) for whch g(x) has the same value. Snce g(x) s nonconstant, ts degree must be greater than Let p(x) be the monc real polynomal of degree n. If n =, then p(r) = r + a for some real number a, and p(x) s the average of x and x + 2a, each of whch has real root. Now we assume that n >. Let g(x) = (x 2)(x 4) (x 2(n )). The degree of g(x) s n. Consder the polynomals q(x) = x n kg(x), r(x) = 2p(x) q(x) = 2p(x) x n + kg(x). 3

4 We wll show that for large enough k these two polynomals have n real roots. Snce they are monc and ther average s clearly p(x), ths wll solve the problem. Consder the values of the polynomal g(x) at n ponts x =, 3, 5,..., 2n. These values alternate n sgn and are at least (snce at most two of the factors have magntude and the others have magntude at least 2). On the other hand, there s a constant c > 0 such that for 0 x n, we have x n < c and 2p(x) x n < c. Take k > c. Then we see that q(x) and r(x) evaluated at n ponts x =, 3, 5,..., 2n alternate n sgn. Thus q(x) and r(x) each has at least n real roots. However snce they are polynomals of degree n, they must have n real roots, as desred. 0. Wthout loss of generalty, suppose n = deg(f) deg(g). Let r,..., r k be the dstnct roots of f and let s,..., s l be the dstnct roots of f. We clam that k + l n +. Indeed, suppose Then Smlarly, f then f(x) = (x r ) p (x r k ) p k. (x r ) p (x r k ) p k f. f(x) = (x s ) q (x s l ) q l, (x s ) q (x s l ) q l f. Snce the roots of f and f are dstnct, (x r ) p (x r k ) p k (x s ) q (x s l ) q l f. Snce f has degree n, (p ) (p k ) + (q ) +... (q l ) n. Snce p p k = q q l = n, ths gves (n k) + (n l) n, or k + l n +. Now f g has degree at most n and has at least n+ dstnct roots r, r 2,..., r k, s,..., s l, so t must be dentcally 0, and f = g.. Suppose f, g have no common nonconstant factor. Consder f, g as elements of C(x)[y]. Snce C(x) s a feld, C(x)[y] s a prncpal deal doman. From Gauss s Lemma appled to the rng C[x] and ts fracton feld C(x), any two polynomals wth no common factor n C[x][y] have no common factor n C(x)[y]. Hence the greatest common dvsor of f, g n C(x)[y] s, and we can wrte = u(x, y)f(x, y) + v(x, y)g(x, y) for some u, v C(x)[y]. Multplyng to clear denomnators n u, v, we get p(x) = s(y)f(x, y) + t(y)g(x, y). Any zero (x, y) of both f and g must satsfy p(x) = 0. Snce p s a nonzero polynomal, t has fntely many zeros,.e. there are fntely many possbltes for x. Each value gves a polynomal equaton n y, whch agan has fntely many zeros. 4

5 (B) Arthmetc Propertes. Take n so that P (n) s nonzero. Suppose t s prme. Now P (n) P (n + kp (n)) P (n) P (n) P (n + kp (n)) for all k Z. If P (x) s not composte for all nteger x, then P (n + kp (n)) s ±P (n) or 0 for all k Z. P attans one of these values nfntely many tmes so s constant. 2. If not, then no two are equal. Wthout loss of generalty, assume that c s between a and b. Then P (a) P (b) = c b < b a. However, b a P (b) P (a), a contradcton. 3. We frst show that every fxed pont x of Q s n fact a fxed pont of P P. Consder the sequence gven by x 0 = x and x + = P (x ) for 0. Assume x k = x 0. Then d := x + x P (x + ) P (x ) = x +2 x + = d + for all, whch together wth d k = d 0 mples d 0 = d = = d k. Suppose d = d 0 = d 0. Then d 2 = d as otherwse x 3 = x and x 0 wll never occur n the sequence agan. Smlarly d = d for all and x = x 0 + d x 0 for all, a contradton. It follows that d = d 0, as clamed. Thus we can assume Q = P P. If every nteger t wth P (P (t)) = t also satsfes P (t) = t, the number of solutons s at most deg(p ) = n. Suppose P (t ) = t 2, P (t 2 ) = t, P (t 3 ) = t 4, P (t 4 ) = t 3, where t s not equal to any of t 2, t 3, t 4. Snce t t 3 dvdes t 2 t 4 and vce versa, we conclude t t 3 = ±(t 2 t 4 ). Assume t t 3 = t 2 t 4,.e. t t 2 = t 3 t 4 = u 0. Snce the relaton t t 4 = ±(t 2 t 3 ) smlarly holds, we obtan t t 3 + u = ±(t t 3 u), whch s mpossble. Therefore, we must have t t 3 = t 4 t 2, whch gves us P (t ) + t = P (t 3 ) + t 3 = c for some c. It follows that all ntegral solutons t of the equaton P (P (t)) satsfy P (t) + t = c, and ther number does not exceed n. 4. Induct on the degree. If f has degree n wth leadng coeffcent c, then the frst nonzero term n the representaton must be c u n p n (x) and f(x) c u n p n (x) can be unquely wrtten as a n p n (x)+...+a p (x)+a 0 p 0 (x) by the nducton hypothess. 5. The assertons (a) (b) and (c) (a) are clear ( ( x ) are ntegers for all ntegers x and nonnegatve ntegers ). Suppose (b) holds. Frst assume that f(x) takes on nteger values at 0,,..., n. We nductvely buld the sequence a 0, a,... so that the polynomal ( ) ( ) ( ) x x x P m (x) = a m + a m + + a 0 m m 0 matches the value of f(x) at x = 0,..., m. Defne a 0 = f(0); once a 0,..., a m have been defned, let a m+ = f(m + ) P m (m + ). Notng that ( x m+) equals at x = m+ and 0 for 0 x m, ths gves Pm+ (x) = f(x) for x = 0,,..., m +. Now P n (x) s a degree n polynomal that agrees wth f(x) at x = 0,,..., n, so they must be the same polynomal. 5

6 Now f f takes on nteger values for any n + consecutve values, then by the argument above on the translated functon t takes on nteger values for all x; n partcular, for x = 0,,..., n. Use the above argument to get the desred representaton n (c). 6. Replacng f(x) by f(x) f(n) as necessary, t suffces to show lcm[, 2,..., deg(f)] f(m) m Z. Lettng d = deg(f) and wrtng f as n Problem 5, the above expresson equals d ( ) a m d ( ) a m lcm(, 2,..., d) = lcm(, 2,..., d). m =0 Each term s an nteger because lcm(, 2,..., d) for all d. 7. Step Suppose P has degree d. Let Q be the polynomal of degree at most d wth Q(x) = q x for 0 x d. Snce the q x are all ntegers, Q has ratonal coeffcents, and there exsts k so that kq has nteger coeffcents. Then m n kq(m) kq(n) for all m, n N 0. Step 2 We show that Q s the desred polynomal. Let x > n be gven. Now =0 kq x kq m (mod x m) for all ntegers m [0, d] Snce kq(x) satsfes these relatons as well, and kq m = kq(m), and hence Now kq x kq(x) (mod x m) for all ntegers m [0, d] kq x kq(x) (mod lcm(x, x,..., x d)). (4) lcm(x, x,..., x ) = lcm[lcm(x, x,..., x ), x ] lcm(x, x,..., x )(x ) = gcd[lcm(x, x,..., x ), (x )] lcm(x, x,..., x )(x ) gcd[x(x ) (x ), (x )] lcm(x, x,..., x )(x ) ( + )! so by nducton lcm(x, x,..., x d) x(x ) (x d). Snce P (x), Q(x) have degree d!(d )!! d, for large enough x (say x > L) we have Q(x) ± x(x ) (x d) kd!(d )!! > P (x). By (4) kq x must dffer by a multple of lcm(x, x,..., x d) from kq(x); hence q x must dffer by a multple of x(x ) (x d) from Q(x), and for x > L we must have kq kd!(d )!! x = kq(x) and q x = Q(x). Now for any y we have kq(y) kq(x) kq x kq y (mod x y) for any x > L. Snce x y can be arbtrarly large, we must have Q(y) = q y, as needed. 6

7 8. For N, let (Defne P 0 (x) =.) P (x) = (x (2k )). k= Step By Problem 4, any polynomal wth nteger coeffcents can be wrtten n the form 0 n c P (x). Step 2 Let a = k=0 2. We clam that 2 a P k (x) for all N and all odd x. For a prme p and n Z, denote by v p (n) the exponent of the hghest power of p dvdng n (by conventon v p (0) = ). For gven odd x let f(α) be the number of values of k (0 k ) where 2 α x 2k. Then v 2 (P (x)) = v 2 (x 2k) = k=0 f(α) snce each k wth 2 α x 2k s counted α tmes n ether sum. Snce any set of 2 α consecutve even ntegers has one dvsble by 2 α, any set of consecutve even ntegers has at least 2 ntegers dvsble by 2 α. Hence α f(α) 2, and α v2 (P (x)) α=0 2 as desred. α Note a 0 = 0, a =, a 2 = 3, a 3 = 4, a 4 = 7, a 5 = 8, and a 0 for 6. Step 3 Next, we clam that f P (x) = 0 n c P (x) has a complete remander sequence then c s odd. (c 0 obvously needs to be odd.) We have 4 P (4k+) P () for any nteger ; hence r(4k + ) r() (mod 4) and r(4k + 3) r(3) (mod 4) for each k. In order for the remander sequence to be complete, we need r() r(3) (mod 4). But notng that a 2 and P (x) 0 (mod 4) for odd x and 2, we have P (3) P () c (P (3) P ()) 2c (mod 4). Hence c s odd. Step 4 Snce for any odd x, P (x) s dvsble by 2 a, f we mod out c by 2 0 a, and delete the terms wth P for 6 (where a 0), we get a polynomal wth the same remander sequence as P. If P (x) gves a complete remander sequence, then c 0 s odd, so there are 2 9 choces for t; c s odd, so there are at most 2 8 choces for c (mod 2 9 ) (a = ); for 2 5 there are at most 2 0 a choces for c (mod 2 0 a ). Hence the number of complete remander sequences s at most α= α = = =2 9. Step Let k be an nteger and p be a prme. Call the ( + )-subsgnature of f modulo p k. (f(0),..., f()) (mod p k ) By Problem 4, each polynomal f (Z/p k Z)[x] can be unquely represented as The followng are clear: f(x) = 0 7 a x, a Z/p k Z (5)

8 (a) The term a x does not affect the value of f(0),..., f( ). (b) p! for 0 < p. (c) p! for p < p 2. (d) p x when p. (e) p 2 x when 2p. We show that for m p, the p km choces for a 0,..., a m gve the p km dstnct m- term sequences modulo p k as m-subsgnatures. Ths s clear for m =. Suppose ths s true for a certan m < p. Gven a sequence (y 0,..., y m ), we can fnd a 0,..., a m n (5) such that f() = y for 0 < m. Next note the term a m x m ranges over all of Z/p k Z for x = m snce p m! mples m! s nvertble modulo p k. Hence we can choose a m to make f(m) = y m (ths does not affect f(0),..., f(m ) by tem (a) above), and all p k(m+) dstnct (m + )-term sequences modulo p k are (m + )-subsgnatures. Hence there are p kp possble p-subsgnatures modulo p k. In partcular, there are p possble sgnatures modulo p and p 2 p-subsgnatures modulo p 2. Step 3 Now, we show there are p 3p 2p-subsgnatures (y 0, y,..., y 2p ) modulo p 2, correspondng to the sequences (a 0 mod p 2,..., a p mod p 2, a p mod p,..., a 2p mod p). (6) Agan, we nduct on the followng statement: there are p 2p p k (p + k)-subsgnatures (y 0, y,..., y p+k ) modulo p 2, correspondng to the coeffcents (a 0 mod p 2,..., a p mod p 2, a p mod p,..., a p+k mod p). (7) Ths s true for k = 0 by Step 2; once t has been establshed for some k < p, note that a p+k x p+k ranges over p resdues modulo p 2 when x = p + k, as p (p + k)!. Hence gven the coeffcents (7) there are exactly p possble values for f(p + k), gvng p 2p p k+ possble (p + k + )-subsgnatures. The clam follows. Step 4 The 2p-subsgnature modulo p 2 completely determnes the sgnature modulo p 2. From Step 3, each such subsgnature corresponds to a sequence as n (6). Snce p 2 x for 2p and p x for p, f(x) = a }{{} x + a }{{} }{{} x + a x }{{} 2p 0 mod p 2 p <2p determned mod p 0 mod p <p determned mod p 2 s determned modulo p 2 for each x. Step 5 If p n then there are p p possble sgnatures modulo p and f p 2 n then there are p 3p possble sgnatures modulo p 2. Hence, snce a sequence modulo p vp(n) for every prme p n unquely determnes the sequence modulo n by the Chnese Remander Theorem, there are at most p p p 3p = p n p 2 n p n p p p 2p p 2 n 8

9 possble sgnatures modulo n. All of these are attanable: t suffces to show that gven a vald sgnature s p modulo p vp(n) for every prme p n, we can fnd a polynomal that has sgnature s p modulo p vp(n) for every prme p n. There exsts a polynomal R p wth sgnature s p ; then f = prme p n s the desred polynomal. q n,q p q 2 q n,q p q 2 (mod p 2 ) R p 0. Step A polynomal P (x) has all terms dvsble by a prme p f and only f x(x ) (x (p )) P (x) n Z/pZ. Call a polynomal p-vald f x(x ) (x (p )) P (x) modulo p. Step 2 A polynomal s fne f and only f t s p-vald for every prme p. Indeed, the forward asserton follows from Step. Conversely, suppose that the polynomal s p-vald for every prme p; then there does not exst a prme p such that x(x ) (x (p )) P (x) n Z/pZ. Gven k N, let p,..., p j be all the prmes dvdng k. Then, for j, there exsts x Z/pZ such that P (x ) 0 (mod p ) for all j. Then any x N wth x x (mod p ) causes P (x ) to not be dvsble by any of p,..., p j and hence relatvely prme to k. By the Chnese Remander Theorem we can fnd nfntely many such x, so P (x) s fne. Step 3 Let p <... < p m be all prmes less than n. In S the proporton of p -vald polynomals modulo p s. Indeed, wrte p p P (x) x(x ) (x (p ))Q(x) + R(x) (mod p ), where R(x) s the remander upon dvdng P (x) by R(x). Q(x) can be any polynomal of degree n p, and for any choce of Q(x), exactly p p out of p p choces for R(x) are possble, namely any polynomal of degree less than p except the zero polynomal. Step 4 The equatons P (x) R (x) (mod p ) unquely determne P (x) modulo p p 2 p m. (Ths s a drect applcaton of the Chnese Remander Theorem.) Snce the proporton p -vald polynomals modulo p s p p, the proporton of polynomals n S that are p -vald for each s ( ) m = p p. Now p p 2 p m n! so there are a fxed number of polynomals correspondng to each polynomal modulo p p m that s p -vald for each. Hence the proporton of fne polynomals s at most ( ) m = p p. (C) Dvsblty, GCD, and Irreducblty. There exst polynomals u, v Q[x] so that uf + vg =. Clearng denomnators we get the desred representaton. Now f uf +vg = k for f, g Z[x] then replacng u by (u mod g) and v by (v mod g), we get u f +v g = k (why?) wth u, v havng degrees less than deg(f), deg(g), respectvely. Note that u, v are unquely determned up to 9

10 a constant multple, snce we need u f k (mod g), and u s determned modulo g up to a constant factor. Lettng c be gcd(a 0,..., a m, b 0,..., b n ), we necessarly have u = ±u/c, and v = ±v/c, gvng that k = ±ck We have uf(n)+vg(n) = k for some u, v Z and nonzero. Hence gcd(f(n), g(n)) k. 3. Let f n denote f terated n tmes. We wll encode the statement of the problem n polynomals. For a polynomal p(x) = c n x n + + c x + c 0, we wll denote by p(f) the functon c n f n (x) + + c f(x) + c 0 x. We are gven (x 3 + 2x 4 + x 4)f = 0 and (x 2009 )f = 0. Snce gcd(x 3 + 2x 2 + x 4, x 2009 ) = x and hence there exst polynomals u(x), v(x) Q[x] for whch u(x)(x 3 +2x 2 +x 4)+v(x)(x 2009 ) = x. Then 0 = (u(x)(x 3 + 2x 2 + x 4)f)(t) + (v(x)(x 2009 )f)(t) = ((x )f)(t) = f(t) t as desred. 4. Frst we prove unqueness. In any nonzero multple of (x + 2)(x + 5) = x 2 + 7x + 0, the coeffcent of the term wth smallest exponent must be dvsble by 0. Any two polynomals both satsfyng the gven condton must have dfference dvsble by (x + 2)(x + 5). Gven any two dstnct polynomals Q and R havng dfference dvsble by x 2 + 7x + 0, one coeffcent must dffer by at least 0, so the coeffcents of Q and R cannot all be n [0, 0). Now we show exstence. Start wth the polynomal n and repeat the followng process. If the smallest term wth coeffcent 0 or greater s a n x n, add x n (x + 2)(x + 5)(x ) = x n (x 3 + 6x 2 + 3x 2 0) to ths polynomal (call ths frng at poston n). Changng the polynomal by a multple of (x + 2)(x + 5) does not change ts value at 2, 5. Furthermore, note the sum of coeffcents stays the same. If at some tme the polynomal has all coeffcents less than 0 we are done. Suppose ths process goes on forever. Consder the sequence of trplets (a n, a n, a n 2 ) where a n s the leadng coeffcent of the polynomal. By our process, n 3 must have fred n the prevous step; postons less than n 23 wll never fre agan; terms x, < n 3 have coeffcents less than 0. Snce there are a fnte number of possbltes for (a n, a n, a n 2 ) (as ther sum s bounded), at some later tme the same trplet must appear twce, say as the coeffcents of x n, x n, x n 2 at step t and as the coeffcents of x m, x m, x m 2 at step t 2 > t. The coeffcents of x n 3,..., have not changed n the ntervenng steps. Then the dfference between the polynomals at these two tmes s (x m n )(a n x 2 + a n x + a n 2 ). Snce x 2 + 7x + 0 must dvde ths polynomal but t has no factor n common wth x m n, t must dvde a n x 2 + a n x + a n 2. Ths means that a n x n + a n x n + a n 2 x n 2 evaluates to 0 at 2, 5. Hence f we remove these terms at tme t, then we get a polynomal satsfyng the desred condtons. 5. By dvson wth remander we can wrte f(x) = q(x)g(x) + r(x) where deg(r) < deg(f). Then f(n) g(n) = q(x) + r(n) g(n). 0

11 When n s large enough, r(n) < g(n). Hence r(n) = 0 for nfntely many n, and n fact r(x) = 0, g f. 6. Let f(x) = x n + x 2 and g(x) = x m + x. By the prevous problem, we need f(x) g(x). Both f, g have a unque root n (0, ), snce both are ncreasng and go from to. Ths must be the same root snce g f; call t α. We use α to show m < 2n. Certanly α > φ where φ s the postve root of h(x) := x 2 x + = 0. Ths s because f s ncreasng n (0, ) and f(φ) < h(φ) = 0 = f(α). On the other hand, f m 2n then α = α m (α n ) 2 = ( α 2 ) 2 and the outer terms rearrange to gve α(α )(α 2 α + ) 0, whch requres α φ, a contradcton. We show that the only soluton wth m < 2n s (m, n) = (5, 3). Snce x n + x 2 (x m n (x n + x 2 ) (x m + x )), x n + x 2 x m n+2 x m n x +. Snce x m n+2 x m n x+ s not dentcally 0, and snce m 2n, m n+2 s ether m or m+. In the frst case, we must have x m n+2 x m n x+ = x n +x 2, whch s mpossble. In the second case, we must have (x )(x n +x 2 ) = x n+ x n x+ so x n = x n + x 3 x 2, whch can only happen f n = 3, and therefore m = 5. As (x 3 + x 2 )(x 2 x + ) = (x 5 + x ), ths soluton works. 7. In lght of Problem, we just need to fnd u, v wth uf + vg a constant, deg(u) < deg(g), and wth the coeffcents of f, g together havng gcd. Ths constant wll be the desred k 0. For polynomals f, g defne (f mod g) to be the remander when f s dvded by g. For an nteger d and f Z we wrte d f f d s the maxmal postve nteger that dvdes every coeffcent of f. We need the followng facts. (a) (from Gauss s Lemma) If p, q Z[x] and d p, e q then de pq. (b) For nteger n, nteger k, 0 < k < 2 n, ( 2 n k ) s even. Moreover, (( ) ( )) 2 n 2 n gcd,..., 2 n = 2. (c) If u, v Z[x], v s monc, there exst q, r Z[x] such that u = qv + r, deg(r) < deg(u). Let P (x) = (xn ) n. Note the roots of x n + are e πk/n for k odd, 0 < k < 2n. For ( 2) n any root x of x n + = 0, P (x) = (xn ) n ( 2) n = ( 2)n ( 2) n =. So P (x) (mod x n + ). Snce ( ) n = 0, x + dvdes x n. Hence (x + ) n (xn ) n or ( 2) n P (x) 0 (mod (x + ) n ). Therefore, for some polynomals f, g, P (x) = (x + ) n f = (x n + )g +.

12 ( x We have f(x) = n n, ( 2)(x+)) and ( 2) n = ( ) x n n (x + ) n + ( 2) n g(x)(x n + ). x + We can wrte ( ) x n n = s(x)(x n + ) + t(x) x + where deg(t) < deg(x n + ) = n, and fnd ( 2) n = t(x)(x + ) n + [( 2) n g(x) + s(x)(x + ) n ](x n + ). Now we fnd the d N such that d t(x). Clam 2 (2r )q s the hghest power of 2 dvdng [( ) x n n x+ mod x 2 r + ]. We have For 0 < r x n x + = (x )(x2 ) (x 2r )(x 2r (q ) + + x + ). 2 r ( 2 r (x 2 ) 2r = k k=0 ) x 2k ( ) k 2 r k= ( 2 r k ) x 2k ( ) k (mod x 2r + ). By tem (c), 2 ( ) 2 r k for k 2 r. Let f (x) = 2 r 2 k= Note that f Z[x] and We have f () = 2 ((x2 ) 2r 2) x= =. ( 2 r k ) x 2 k ( ) k. (x 2 ) 2rq = [(x 2 ) 2r ] 2q 2 2q f (x) 2 q (mod x 2r + ) Multplyng these equatons for 0 < r we get r [(x )(x 2 ) (x 2r )] n = 2 ( r )q f (x) 2 q =0 r = 2 (2r )q f (x) 2 q =0 (mod x 2r + ) [( r ) ] Let w(x) = =0 f (x) 2 q (x 2r (q ) + + x + ) n. By tem 3, we can wrte, w(x) = w (x)(x 2r + ) + w 2 (x); w, w 2 Z[x]. Then w() s odd so the sum of coeffcents of w(x) s odd and not all coeffcents of w 2 (x) are even. Hence ( ) x n n mod x 2r + = [(x )(x 2 ) (x 2r )(x 2r (q ) + + x + )] n mod x 2r + x + = 2 (2r )q [w(x) mod x 2r + ] 2

13 and 2 (2r )q s the hghest power of 2 dvdng xn mod x+ x2r +. Clam 2 2 2rq s the hghest power of 2 dvdng ( ) x n n x+ mod x 2 r (q ) x 2r +. We have x 2r (q ) + + x 2r + 2x 2r (mod x 2r (q ) x 2r + ) so ( ) x n n x+ s congruent to 0<<q, odd [(x )(x 2 ) (x 2r )(x 2r (q ) + + x 2r + )] n ( n [(x )(x 2 ) (x 2r )] n 2 n x ) 2r (mod x 2r (q ) x 2r + ) 0<<q, odd and the clam follows. Let R = ( ) x n n x+ mod x 2 r + and R 2 = ( ) x n n x+ mod x 2 r (q ) x 2r +. Let Q = 0<<q, odd x2r, P = <q x2r, P 2 = x 2r +. Then P = QP 2 + R R 2 = (R R 2 )(P qp 2 ) R P (R R 2 ) = R 2 P 2 Q(R R 2 ) Let F = R P (R R 2 ). Then F R (mod x 2r +) and F R 2 (mod x 2r (q ) x 2r + ). Hence by the Chnese Remander Theorem, snce t(x) R (mod x 2r + ), t(x) R 2 (mod x 2r (q ) x 2r + ), and deg(f ) < n we have F (x) = t(x). Let G(x) = ( 2) n g(x) s(x)(x + ) n. Then ( 2) n = F (x)(x + ) n + G(x)(x n + ). Note 2 (2r )q s the hghest power of 2 dvdng R, and 2 2rq R 2, F (x) = R (x) P (x)(r (x) R 2 (x)) = ( P (x))r (x) + P (x)r 2 (x). 2 P (x) so 2 (2r )q s the hghest power of 2 dvdng S (x) = ( P (x))r (x). There exsts j so that the coeffcent u j of x j n S (x) s not dvsble by 2 (2r )q+. Let the coeffcent of x j n P (x)r 2 (x) be b j. Note 2 2rq R 2 mples 2 2rq b j. Hence 2 (2r )q+ a j + b j, and 2 (2r )q+ F (x). Let F(x) = F (x) 2 (2r )q and G(x) = G(x) 2 (2r )q. Then F(x) Z[x]. Let F(x) = l =0 c x, G(x) = m =0 d x. Then 2 q = 2 2r q (2 r )q = F(x)(x + ) n + G(x)(x n + ). Snce (x n +) 2 q F(x)(x+) n n Q[x], by dvsblty holds n Z[x],.e. G(x) Z[x]. 2 gcd(c 0,..., c l ) and gcd(c 0,..., c l, d 0,..., d m ) 2 q. Hence the gcd s. By Problem, k 0 = 2 q. (D) Algebrac Numbers. Let β be the root of greatest absolute value. Suppose by way of contradcton f(α 3 + ) = 0. Snce f(x) s rreducble and f(α) = 0, t s the mnmal polynomal of α. Snce the polynomal g(x) = f(x 3 + ) has α as a root f(x) f(x 3 + ). Hence f(β) = f(β 3 + ) = 0. Let r = β. Now r 3 r > 0 for r 3 2, so β 3 + β 3 > β, contradctng the maxmalty of β. 3

14 2. Suppose the LHS s not 0. The conjugates of c a c n a n are among the numbers c b c n b n where b s a conjugate of a. Each of these have absolute value at most c a + + c n a n. The product of all conjugates (ncludng c a c n a n ) s an nteger because t s the constant term of the mnmal polynomal of c a c n a n ; hence t s at least. The conjugates multply to an nteger wth absolute value at least, they number at most d d n and each s at most ( ) d c a + + c n a d 2 d n n ; hence each s at least c a. + + cna n 3. Ths tme we can do better than n Problem 2. Suppose ω,..., ω k are the roots of unty. We clam that p (x (ω ωk)) (8) = has nteger coeffcents snce each coeffcent. Indeed, we can wrte each coeffcent n the form a 0 + a ω + + a p 2 ω p 2, (9) where ω s a prmtve pth root of unty. If we replace ω by ω for some p, then the product n (8) stays the same. Hence (9) stays the same as well, and snce, ω,..., ω p 2 satsfy no lnear dependency relaton, we must have a 0 = = a p 2, and (9) s an nteger. (A more advanced way of lookng at ths s the followng: the coeffcents are n the fxed feld of all automorphsms of Q(ω) fxng Q; Galos Theory allows us to conclude that these coeffcents are actually n Q.) Thus the conjugates of ω + + ω k are among the ω ωk. Snce each has absolute value at most k, and ther product s a nonzero nteger (and hence has absolute value at least ), each must be at least. k p 2 (E) Cyclotomc and Chebyshev Polynomals. Let the sde lengths be a 0,..., a p. Then lettng ω be the pth root of unty, a p ω p + + a 0 = 0 Hence ω satsfes the polynomal equaton a p x p + + a 0. Snce the mnmal polynomal of ω s Φ p = x p + + x +, x p + + x + a p x p + + a 0,.e. a p = = a The factorzaton of x n nto rreducble (monc) factors P P k gves a (not necessarly complete) factorzaton of 2 n. If P has degree d then P (2) 3 d snce P (2) = (2 ω) ω root of P and all roots have absolute value. Lettng ω be a prmtve nth root of unty, the rreducble factor contanng ω s Φ n/gcd(n,), whch has degree ϕ(m) ϕ(n) for some factor m of n. Hence t suffces to fnd n so that 3 ϕ(n) 2 n 993, or ϕ(n) 4 n ln(2) 993 ln(3).

15 However, for any C > 0 we can fnd n so that ϕ(n) < Cn. The product p prme s nfnte (rewrte as geometrc seres, expand to get the harmonc seres). Hence p p prme = 0, and takng n = p p p k for suffcently many dstnct prmes p we fnd that we can make ϕ(n)/n = k = as close to 0 as desred. 3. (A) Let ω = cos(pπ) + sn(pπ). Then ω s a root of unty and hence an algebrac nteger; smlarly ω s as well. Hence 2 cos(pπ) = ω + ω s an algebrac nteger; snce t s ratonal t s a ratonal nteger. Hence we must have 2 cos(pπ) = 0, ±, ±2, gvng p = 0, 3, 2, 2 3,. (B) Step If a s nonreal, aa Q and the rreducble polynomal of a over Q has degree n, then the rreducble polynomal of a + a has degree n/2. Indeed, [Q(a) : Q(a + a)] = 2 snce a satsfes a quadratc equaton n Q[a + a] (namely x 2 (a + a)x + aa = 0), and a R mples a Q(a + a). Snce p p n = [Q(a) : Q] = [Q(a) : Q(a + a)][q(a + a) : Q], the desred result follows. In partcular, ths holds for roots of unty. Step 2 If cos(pπ) = (ω+ω) has degree n over Q, then ω has degree 2n. However, f 2 2p = q n reduced form, then ω s a rth root of unty, and ts rreducble polynomal r s the cyclotomc polynomal Φ r. Hence 2n = ϕ(r). If n = 2, then r = 4, and the solutons to ϕ(r) = 4 are 5, 0, 2. Hence n addton to the p above, ratonal p wth denomnators 5 or 6 also work. Try to use ths method to descrbe the factorzaton of Chebyshev polynomals! 4. It s easy to check that n + + n s a conjugate of n + n. If cos pπ = 2 ( n + n) then 2 ( n + n) s a zero of some Chebyshev polynomal. But then 2 ( n + + n) > s also a zero, a contradcton snce all zeros of Chebyshev polynomals are n [, ]. 5. Soluton Let cos(kθ) = p Q, cos[(k + )θ] = q Q. Then cos θ = cos[(k + )θ] cos θ ± sn[(k + )θ] sn kθ = pq ± ( p 2 )( q 2 ) = a ± b for some a, b Q such that b s not a perfect square. Now T k (a± b) = T k (cos(θ)) = cos(kθ) = p so ether a + b or a b s a zero of the polynomal T k (x) p Q[x]. The mnmal polynomal n Q[x] havng a + b (or a b) as a root s R(x) = [x (a + b)][x (a b)] so R(x) T k (x) p. Now f a b [, ] then we have T k (a b) [, ] and hence not equal to p, a contradcton. So there must exst φ so that cos(φ) = a b. By smlar reasonng a b s a root of T k+ (x) q = 0. 5 p p

16 Now cos(kφ) = T k (a b) = p and smlarly cos[(k + )φ] = q. In summary, cos θ = a ± b (0) cos φ = a b () cos kθ = cos kφ = p (2) cos[(k + )θ] = cos[(k + )φ] = q (3) Now (2) mples kφ = 2πj ± kθ for some j Z, or and (3) mples φ = 2πj k ± θ (4) φ = 2πl (k + ) ± θ (5) for some l Z. Snce b 0, a + b a b and cos θ cos φ. If (4) or (6) hold true for the same sgn, then 2πj = 2πl and j(k + ) = lk; snce gcd(k, k + ) = k k+ but k j ths s a contradcton. So equatng (4) and (6) gves θ = 2π m 2k(k + ) for some m Z. Snce cos(θ) s the root of a quadratc, we know by Problem 3 that cos θ = 3/2, /2, ( 5 ± )/2. It s easy to check that θ = π/6 s the only soluton. Soluton 2 We clam that f cos kθ and cos lθ are ratonal for relatvely prme postve ntegers k, l, then ether cos θ s ratonal or θ s a ratonal multple of π. Let cos kθ = p and cos lθ = q as before. Notng that e φ + that e θ s a root of e φ = 2 cos φ, we have x k + x k = 2p = x2k 2px k + = 0 (6) x l + x l = 2q = x 2l 2qx l + = 0 (7) 2πj ±θ+ Suppose θ s not a ratonal multple of π. Then the numbers e k for 0 j < k 2πj ±θ+ are all dstnct. They are the roots of (6). Smlarly, e l are the roots of (7). Snce θ s not a ratonal multple of π and k, l are relatvely prme, we have that the only soluton to e ±θ+ 2πj k = e ±θ+ 2πj 2 k for 0 j < k and 0 j 2 < l for some choces of sgns s when j = j 2 = 0. Thus the only roots the two polynomals share n common are ω = e θ and ω = e θ, and hence the polynomal g(x) := (x e θ )(x e θ ) s the greatest common dvsor of polynomals x 2k 2px k + and x 2l 2qx l +, whch have ratonal coeffcents. Hence g(x) has ratonal coeffcents, and e θ +e θ = 2 cos θ s ratonal. Now by Problem 3, the only ratonal multples of π that gve a ratonal values for cosne are multples of π. By our clam, θ s a ratonal multple of π. Hence kθ and 6 (k + )θ are both multples of π, and so must θ. Snce cos θ s rratonal, θ = π

17 6. Let p(x) = x n + a n x n + + a 0. Snce x [, ], we set x = cos(θ). Step Wrte p(cos θ) n the form p(cos θ) = b n cos(nθ) + b n cos((n )θ) + + b 0. We know that cos nθ = T n (cos θ), so ths s possble by Problem B4. Moreover, snce the leadng coeffcent of T n s 2 n (by nducton), we must have b n = 2 n. Step 2 The maxmum of h(cos θ) = c n cos(nθ) + + c cos θ + c 0 s at least c n. Assume WLOG c n > 0. Suppose by way of contradcton that max h(cos θ) < c n. Then ( )) ( ( )) ( ( )) kπ kπ kπ c n T n (cos h cos = ( ) k c n h cos n n n s postve for even k and negatve for odd k n the range 0 k n. We conclude that c n T n (x) h (x) has at least n roots n [0, ]. However ths polynomal has degree less than n, contradcton. Applyng Step 2 to p gves the desred result. 7. By nducton, the constant term of T k s 0 for k odd and ± for k even, and the leadng coeffcent s 2 k. Hence by Veta s formula, the product of the roots of T 2n s Snce cos π 4n cos π 4n 3π (2n )π cos cos 4n 4n cos π 4n (F) Polynomals n Number Theory 3π (4n )π cos cos 4n 4n = ± cos (2n+)π 4n 3π (2n )π cos cos 4n 4n = ± 2 2n. cos (2n+3)π 4n =. 2 n 2. By Fermat s Lttle Theorem, x p (mod p). Thus n Z/pZ, cos (4n )π, we get 4n p x p (x ) (mod p). (8) = Wrte (x ) p = p =0 a x. Then matchng coeffcents on both sdes of (8) gves a 0 (mod p) for all < p. (9) Snce p 5, lettng x = p gves (p )! = (x ) p = p p + ( p 2 ) a p + a p + (p )! snce ( )( 2) ( p + ) = ( ) p (p )! = (p )!. Subtractng (p )! on both sdes, ( p ) 0 = p p + a p + a p. 7 j=2 =2

18 Usng (9), p 3 a p for 2 p. Hence, snce p 5, p 3 p p + p =2 a p. Snce p 3 dvdes the LHS, p 3 a p and p 2 a. Now p 3 (kp) p + ( p 2 =2 a (kp) ) as well and we get Now, (kp ) p = (x ) p x=pk = (kp) p + ( p ) a (kp) + a p + (p )! j=2 (p )! (mod p 3 ). (20) ( ) pa = (pa)pb pb (pb)! a =a b+ = [(p)(p )p ] b = [(p)(p )p ] [ b ] = ab [p( + a b) ] p b! (p ) p = [ b ] = ab [p( + a b) ] p b! (p ) p By (20), [p( + a b) ] p (p ) p (mod p 3 ). modulo p 3, as needed. 2. Contnung the above, we have by Vete s formula that ( a = (p )! ). p We ve shown that p 2 a, as needed. 3. Rearrangng, t suffces to show (p 2 ) p (p )! (mod p 4 ) = (2) Hence (2) becomes ( ) a b Contnung the above, f we plug n p 2 nto the polynomal (x ) p we get p 2 (p 2 ) p = p 2(p ) + a p 2 + a p 2 + (p )! (p )! (mod p 4 ). 4. Consder the polynomal =2 P (x) = (x + a)(x + b)(x + c) (x d)(x e)(x f) = Sx 2 + Qx + R where Q = ab + bc + ca de ef fd and R = abc + def. Snce S Q, R, t follows that S P (x) for every x Z. Hence S P (d) = (d + a)(d + b)(d + c). Snce S > d + a, d + b, d + c and thus cannot dvde any of them, t follows that S must be composte. 8

19 5. Note that for any odd p, p n! + has a fnte number of postve nteger solutons because for n p, p n!, p n! +. Suppose the solutons to n! + 0 (mod p) are exactly n < n 2 < < n m wth m >. Then for each, < m we have Dvdng these 2 relatons gves n +! (mod p) n! (mod p) (n + )(n + 2) (n + ) (mod p). Lettng k = n + n, we see that x = n s a soluton to For each k, k < p the polynomal (x + )(x + 2) (x + k) (mod p). (x + )(x + 2) (x + k) Z/pZ[x] has at most k roots modulo p. Therefore, there are at most k ndces where n + n = k, for each k wth k < p. Now let j be the smallest postve nteger such that m j(j+) 2. Then (j+)(j+2) 2 m j(j+) 2 and m p > n m n = (n + n ) = = j = 2 j(j + )(2j + ) 6 where the second nequalty follows from the fact that for a fxed k, k < p, n + n has at most k solutons. Snce m j(j+) = j 2 = j, when the dfferences n + n are wrtten n ascendng order, the frst s at least, the next two are at least 2, and so on, each tme the next dfferences are at least and hence sum to at least 2, up to = j. Now snce lm j j(j + )(2j + )/6 ((j + )(j + 2)/2) 3/2 s fnte and greater than 0, there exsts a constant 0 < c < such that ths expresson s greater than c for each j N; lettng c = c we see that p > j(j + )(2j + ) 6 ( ) 3/2 (j + )(j + 2) c c m 3/2 2 or m < cp 2/3. (Requrng c < leads to cp 2/3 > so ths s true even f m = 0,.) 9

20 6. We contnue from Problems -3. The numerator of f p (x) f p (y) n lowest terms s dvsble by p 3 ff p k= p (px + k) 2 k= (py + ) 2 0 (mod p3 ), where the nverse s taken modulo p 3. Frst note that snce p a j = ( ) p j (p )! < < j p 2 j for j p by Veta s formula, < < j p 2 j 0 (mod p). From Problem 2, Now <j<k p Hence by Veta s, ( jk = p a 3 = (p )! Let P (x) = (x ) p. Wrte Then p = 0 (mod p2 ). ) 3 ( p ( <j<k p ) ( 3 3 <j p ) 0 (mod p 2 ). jk p f(t) := [t (px + )] = t p + b p 2 t p b t + b 0. = ) ( j p ). t p + b p 2 t p b 0 = (t px) p + a p 2 (t px) p a (t px) + a 0. Matchng the coeffcent of t 2 on both sdes, usng the Bnomal Theorem, p ( ) n b n a n (px) n 2 2 n=2 a 2 + a 3 p + a 4 p 2 a 2 (mod p 3 ) 20

21 (We used p > 5 p a 4.) By Veta s formula, ( b 2 = (px + p ) p By Wolstenholme, <j p ). (px + )(px + j) (px ) p (py ) p (mod p 3 ). Usng Wolstenholme and a 2 b 2 (mod p 3 ), we get ( ) (px + )(px + j) <j p <j p (mod p 3 ). So p k= ( p (px + k) 2 k= 0 2 ( p k= p k= ) 2 2 (px + k) 2 <j p (py + k) 2 <j p (px + )(px + j) ) 2 2 (py + k) 2 (mod p 3 ) <j p (px + )(px + j) (py + )(py + j) 2