“物不知数问题”的一般化2019-08-29 21:35·初等数学学习aoe1981“物不知数问题”的一般化2019年8月29日星期四本文接前文:——《用现代数学方法解古题“物不知数”》——《用“辗转相除法”将两数的最大公因数表成两数的线性组合》——《完整例解增强版“物不知数”》——《除数不满足“两两互素”条件的“物不知数问题”初探》对于“物不知数问题”的一般化推广,本文打算分两步:一、形如:ax≡b(mod m)或ax-b≡0(mod m)的一元一次同余方程组;二、形如:f(x)≡b(mod m)或f(x)-b≡0(mod m)的一元同余方程组。且以第一步为主介绍,第二步推广据说连数学家都没有找到好的求通解的方法。文中图片均来自网络一、f(x)=a1x+a0数学符号总让人抓狂,且看古题改编:“今有物,不知其数。二倍之,十五、十五数之,剩十四;三倍之,十二、十二数之,剩三;四倍之,十四、十四数之,剩八;五倍之,九九数之,剩五。问:物几何?”数学化:方程组A:2x≡14(mod 15) 式①3x≡3(mod 12) 式②4x≡8(mod 14) 式③5x≡5(mod 9) 式④或者:2x-14≡0(mod 15)3x-3≡0(mod 12)4x-8≡0(mod 14)5x-5≡0(mod 9)这道题也是我精心设计的,因为它一定有解。胡乱构造一道这样的题,无解的可能性是很大的。怎么解?似乎又有了新变化、新挑战,的确如此。步骤1:运用定理:若:x≡b(mod m), 则:kx≡kb(mod km)。 其中:x、b、m、k∈Z。由于:[2,3,4,5]=60原方程组变为:60x≡420(mod 450),取k=30;60x≡60(mod 240),取k=20;60x≡120(mod 210),取k=15;60x≡60(mod 108),取k=12。步骤2:令:y=60x则有:方程组B:y≡420(mod 450) 式①y≡60(mod 240) 式②y≡120(mod 210) 式③y≡60(mod 108) 式④步骤3:判断方程组B是否有解。先对模进行标准素因子分解:m1=450=2×3^2×5^2m2=240=2^4×3×5m3=210=2×3×5×7m4=108=2^2×3^3判断:(m1,m2)|(b2-b1)=(450,240)|(60-420)=30|(-360)(m1,m3)|(b3-b1)=(450,210)|(120-420)=30|(-300)(m1,m4)|(b4-b1)=(450,108)|(60-420)=18|(-360)(m2,m3)|(b3-b2)=(240,210)|(120-60)=30|60(m2,m4)|(b4-b2)=(240,108)|(60-60)=12|0(m3,m4)|(b4-b3)=(210,108)|(60-120)=6|(-60)结论:方程组B有解。步骤4:拆解方程组B的合数模。y≡420(mod 2)y≡420(mod 9)y≡420(mod 25)——————y≡60(mod 16)y≡60(mod 3)y≡60(mod 5)——————y≡120(mod 2)y≡120(mod 3)y≡120(mod 5)y≡120(mod 7)——————y≡60(mod 4)y≡60(mod 27)将常数项继续模相应m运算得:y≡0(mod 2)y≡6(mod 9)y≡20(mod 25)——————y≡12(mod 16)y≡0(mod 3)y≡0(mod 5)——————y≡0(mod 2)y≡0(mod 3)y≡0(mod 5)y≡1(mod 7)——————y≡0(mod 4)y≡6(mod 27)去重:y≡0(mod 2)y≡6(mod 9)y≡20(mod 25)——————y≡12(mod 16)y≡0(mod 3)y≡0(mod 5)——————y≡1(mod 7)——————y≡0(mod 4)y≡6(mod 27)整理:y≡0(mod 2)y≡0(mod 4)y≡12(mod 16)——————y≡0(mod 3)y≡6(mod 9)y≡6(mod 27)——————y≡0(mod 5)y≡20(mod 25)——————y≡1(mod 7)保留高次幂模:y≡12(mod 16) 式①y≡6(mod 27) 式②y≡20(mod 25) 式③y≡1(mod 7) 式④此时,模{16,27,25,7}满足“两两互素”的条件,是为方程组C。从方程组A到B再到C,一路变形,皆恪守“等价变形”,否则就误入歧途了。步骤5:求方程组C的特解参数:v1、v2、v3、v4。v1:{m1,m2m3m4}={16,27×25×7}={16,4725}1=886×16-3×4725(辗转相除过程略)v1=-3——————v2:{m2,m1m3m4}={27,16×25×7}={27,2800}1=-1037×27+10×2800(辗转相除过程略)v2=10——————v3:{m3,m1m2m4}={25,16×27×7}={25,3024}1=121×25-1×3024(辗转相除过程略)v3=-1——————v4:{m4,m1m2m3}={7,16×27×25}={7,10800}1=1543×7-1×10800(辗转相除过程略)v4=-1步骤6:代入特解模型求出特解。c=v1(m2m3m4)b1+v2(m1m3m4)b2+v3(m1m2m4)b3+v4(m1m2m3)b4=-3×4725×12+10×2800×6+(-1)×3024×20+(-1)×10800×1=-170100+168000-60480-10800=-73380求通解。y=c+k[m1,m2,m3,m4]=-73380+k×[16,27,25,7]=-73380+75600kx=y/60=(-73380+75600k)÷60=-1223+1260k当k=1时,得最小正数解:x=37。步骤7:验证。2×37÷15=4……143×37÷12=9……34×37÷14=10……85×37÷9=20……5结论:形如:f(x)≡b(mod m)的一元同余方程组中,当f(x)=a1x+a0时,可以完整解决。二、f(x)=anx^n+……+a3x^3+a2x^2+a1x+a0此时的一元同余方程组主要是指一元高次同余方程组,形如:f(x)≡b1(mod m1)g(x)≡b2(mod m2)h(x)≡b3(mod m3)……其中:f(x)、g(x)、h(x)……全是对于不定量(或变量、未知数)x的多项式,且彼此间次数不见得一致。此时的求解一元同余方程组并不容易,甚至据说连一元二次同余方程组的通解都没能找到……要是再推广的话,可以有二元同余方程组、三元同余方程组、甚至更多……f(x,y,z)≡b1(mod m1)g(x,y,z)≡b2(mod m2)h(x,y,z)≡b3(mod m3)……实非寻常人可为啊。不过,对于一元同余方程组,哪怕是高次的,却是有一种“暴力”求解方法的,容后文介绍。
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去(蚕耻)年(狈颈补苍)4月(驰耻别)马(惭补)斯(厂颈)克(碍别)成(颁丑别苍驳)为(奥别颈)罢飞颈迟迟别谤董(顿辞苍驳)事(厂丑颈)会(贬耻颈)成(颁丑别苍驳)员(驰耻补苍),当(顿补苍驳)时(厂丑颈)他(罢补)已(驰颈)经(闯颈苍驳)同(罢辞苍驳)意(驰颈)收(厂丑辞耻)购(骋辞耻)罢飞颈迟迟别谤,多(顿耻辞)西(齿颈)说(厂丑耻辞),在(窜补颈)此(颁颈)之(窜丑颈)前(蚕颈补苍)他(罢补)曾(窜别苍驳)多(顿耻辞)次(颁颈)邀(驰补辞)请(蚕颈苍驳)马(惭补)斯(厂颈)克(碍别)加(闯颈补)入(搁耻)罢飞颈迟迟别谤董(顿辞苍驳)事(厂丑颈)会(贬耻颈)。多(顿耻辞)西(齿颈)称(颁丑别苍驳):“马(惭补)斯(厂颈)克(碍别)是(厂丑颈)我(奥辞)们(惭别苍)的(顿别)头(罢辞耻)号(贬补辞)用(驰辞苍驳)户(贬耻),也(驰别)是(厂丑颈)我(奥辞)们(惭别苍)的(顿别)头(罢辞耻)号(贬补辞)客(碍别)户(贬耻),他(罢补)对(顿耻颈)平(笔颈苍驳)台(罢补颈)有(驰辞耻)着(窜丑耻辞)深(厂丑别苍)入(搁耻)的(顿别)理(尝颈)解(闯颈别),而(贰谤)且(蚕颈别)他(罢补)是(厂丑颈)技(闯颈)术(厂丑耻)专(窜丑耻补苍)家(闯颈补),知(窜丑颈)道(顿补辞)如(搁耻)何(贬别)打(顿补)磨(惭辞)技(闯颈)术(厂丑耻)。”
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