r(x)?0or0??r(x)??h(x)
于是f(x)?g(x)?h(x)[q1(x)?q2(x)]?h(x)[f(x)?g(x)]
5.常数a,b,c满足什么条件时,g(x)?x2?ax?1能整除f(x)?x4?bx?c? 解：f(x)?g(x)(x2?ax?a2?1)?(b?2a?a2)x?c?1?a2 所以b?2a?a2?0,c?1?a2?0所以2a?b?c?1?a
6.设f(x),g(x),h(x),q(x)?F[x],h(x)?0,并且满足f(x)q(x)g(x)h(x,)证明:若2h(x)f(x),那么q(x)g(x). 证明：设g(x)h(x)?f(x)q(x)p1(x),f(x)?h(x)p2(x) 所以g(x)h(x)?h(x)p2(x)q(x)p1(x)?g(x)?q(x)p1(x)p2(x)
?q(x)g(x)
7.证明：对任意非负整数n,都有x2?x?1|xn?2?(x?1)2n?1 证明：对n用数学归纳法: 当n?0时,结论显然成立;
假设结论在一切不大于n的非负整数成立,那么在n?1时,
xn?3?(x?1)2n?3?x[xn?2?(x?1)2n?1]?(x?1)2n?1[(x?1)2?x] ?x[xn?2?(x?1)2n?1]?(x?1)2n?1(x2?x?1)
由归纳假设有x?x?1|x所以x?x?1|x2n?32n?2?(x?1)2n?1,同时x2?x?1|(x?1)2n?1(x2?x?1)
?(x?1)2n?3
k8.设k是任意正整数,证明x|f(x)?x|f(x) 证明：充分性是显然的,下面证明必要性
用反证法:若x|f(x),则f(x)?xf1(x)?c,c?0,那么
fk(x)?[xf1(x)?c]k?xg(x)?ck,由x|fk(x)?x|ck矛盾.
9.证明：x|f(x)?f(x)的常数项为0 证明：设f(x)?anx?an?1xnn?1???a1x?a0,an?0于是由于
x|anxn?an?1xn?1???a1x,x|f(x)?anxn?an?1xn?1???a1x?a0
所以x|f(x)?(anxn?an?1xn?1???a1x)?x|a0?a0?0 反过来,若a0?0,显然有x|f(x) 10.证明：xd?1|xn?1?d|n
证明：(?)设n?dq,则xn?1?xdq?1?(xd)q?1?(xd?1)[(xd)q?1?(xd)q?2???1]
?xd?1|xn?1
(?)若d|n,设n?dq?r,0?r?d,于是
xn?1?xdq?r?1?xdqxr?xr?xr?1?xr(xdq?1)?(xr?1)
由于xd?1|xn?1,xd?1|xr[(xd)q?1]?xd?1|xr?1,但0?r?d,这显然不可能.所以,必然有r?0,即d|n. 1.4最大公因式 1.求(f(x),g(x))
(1)f(x)?2x?3x?4x?1,g(x)?x?2x?3 (2) f(x)?x?6x?4x?5,g(x)?x?x?30 (3) f(x)?2x?x?x?x?1,g(x)?x?x?x?1
(4) f(x)?x?5x?7x?2x?4x?8,g(x)?5x?20x?21x?4x?4 解：(1)f(x)?g(x)(2x?3)?4(x?x?2),g(x)?(x?x?2)(x?1)?5x?5
225432432543243322433r1(x)?x2?x?2?(x?1)(x?2)
所以(f(x),g(x))?x?1;
(2) f(x)?g(x)(x?7)?41(x?5),g(x)?(x?5)(x?6) 所以(f(x),g(x))?x?5; (3) (f(x),g(x))?1
(4)f(x)?(x?2)(x?x?1),g(x)?(x?2)(5x?1)所以(f(x),g(x))?(x?2)
322222.设f(x)?xn?a1xn?1???an?1x?ang(x)?xn?1?a1xn?2???an?2x?an?1,n?1,求(f(x),g(x)) 解： 由于
?g(x)an?0 xg(x)?f(x)?an?f(x)?g(x)(?x)?an?(f(x),g(x))??1a?0n?3.设f(x)?xm?n?xm?xn?1,g(x)?xm?xm?2?2,m?n ,求(f(x),g(x)) 解：
4.对下列各题的f(x),g(x),求u(x),v(x),使(f(x),g(x))?f(x)u(x)?g(x)v(x) （1）f(x)?4x4?2x3?15x2?6x?3，g(x)?2x3?5x?3 （2）f(x)?x4?2x3?x2?4x?2,g(x)?x4?x3?x2?2x?2 （3）f(x)?x4?x3?4x2?4x?2,g(x)?x3?x2?x?1
解（1）f(x)?g(x)(2x?1)?5(x2?x)，g(x)?(x2?x)(2x?2)?3(x?1)
x2?x?x(x?1)所以(f(x),g(x))?x?1
15(x?1)?[g(x)(2x?1)?f(x)])(2x?2)?5f(x)
所以x?1???2?21??2?4x??f(x)??x2?x??g(x)
15?55??15?15u(x)??224221x?，v(x)?x?x? 15151555332（2）f(x)?g(x)?x?2x，g(x)?(x?2x)(x?1)?x?2
x3?2x?(x2?2)x，所以(f(x),g(x))?x2?2
x2?2?g(x)?(x3?2x)(x?1)?g(x)?[f(x)?g(x)](x?1)?(?x?1)f(x)?(x?2)g(x)所以：u(x)??x?1，v(x)?x?2
（3）f(x)?g(x)(x?2)?(x?3x?4)，g(x)?(x?3x?4)(x?4)?7x?17
2249(x2?3x?4)?(7x?17)(7x?4)?128，所以(f(x),g(x))?1
128?49(x2?3x?4)?(17?17)(7x?4)7x?17?g(x)?(x2?3x?4)(x?4)x2?3x?4?g(x)(x?2)?f(x)7x?17?g(x)(9?2x?x)?(x?4)f(x)128?49[g(x)(x?2)?f(x)]?[g(x)(9?2x?x2)?(x?4)f(x)]?(?x?53)f(x)?(x2?51x?107)g(x)?x?53x2?51x?107,v(x)?所以u(x)? 49495.令
2
f(x)与g(x)是F[x]中的多项式，而a,b,c,d是
F中的数，并且满足
ad?bc?0，证明(af(x)?bg(x),cf(x)?dg(x))?(f(x),g(x))
证明：设d1(x)?(f(x),g(x)),d2(x)?(af(x)?bg(x),cf(x)?dg(x)) v(x)?cf(x)?dg(x)) bv(x)?bcf(x)?bdg(x))
令 u(x)?af(x)?bg(x),那么 du(x)?adf(x)?bdg(x),两式相减整理得：
dbf(x)?u(x)?v(x)
ad?bcad?bc同理：cu(x)?acf(x)?bcg(x),av(x)?acf(x)?adg(x))
g(x)?cau(x)?v(x)
bc?adbc?ad由于 d1(x)?(f(x),g(x))?d1(x)|f(x),d1(x)|g(x)
?d1(x)|af(x)?bg(x),d1(x)|cf(x)?dg(x) ?d1(x)|d2(x)
反过来：d2(x)?(af(x)?bg(x),cf(x)?dg(x))?(u(x),v(x))
?d2(x)|u(x),d2(x)|v(x)
db?d(x)|u(x)?v(x)?f(x)2??ad?bcad?bc??
ca?d(x)|u(x)?v(x)?g(x)2?bc?adbc?ad??d2(x)|d1(x)
所以 d1(x)?d2(x)
6.证明定理1.4.7的逆：若(f(x),g(x)h(x))?1，那么(f(x),g(x))?1与
(f(x),h(x))?1都成立。
证明：因为(f(x),g(x)h(x))?1，所以有u(x),v(x)使f(x)u(x)?g(x)h(x)v(x)?1 于是由 f(x)u(x)?g(x)(h(x)v(x))?1?(f(x),g(x))?1 由 f(x)u(x)?h(x)(g(x)v(x))?1?(f(x),h(x))?1 7.设f(x)?0,g(x)?0,证明:
(1)若对任意多项式h(x),由f(x)|g(x)h(x)均可得到f(x)|h(x), 则必有
(f(x),g(x?)1)
(2)若对任意多项式h(x),由f(x)|h(x),g(x)|h(x)均可得到f(x)g(x)|h(x), 则必有
(f(x),g(x))?1
证明：（1）设d(x)?(f(x),g(x))，
那么有 f(x)?d(x)f1(x),g(x)?d(x)g1(x),(f1(x),g1(x))?1 对于 d(x)f1(x)g1(x)?f(x)g1(x)?g(x)f1(x)有f(x)|f(x)g1(x)进而有
f(x)|g(x)f1(x)，取h(x)?f1(x)，依题意有f(x)|f1(x)，但同时有f1(x)|f(x)
所以f(x)?cf1(x),c?0?d(x)?1，即(f(x),g(x))?1 （2）设d(x)?(f(x),g(x))，
那么有 f(x)?d(x)f1(x),g(x)?d(x)g1(x),(f1(x),g1(x))?1 对于 d(x)f1(x)g1(x)?f(x)g1(x)?g(x)f1(x) 所以 f(x)|d(x)f1(x)g1(x),g(x)|d(x)f1(x)g1(x) 取 h(x)?d(x)f1(x)g1(x)