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[提问] 什么情况下是整式?

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发表于 2023-12-5 10:55:18 | 显示全部楼层 |阅读模式

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m和n都是正整数,记\(\D f(a,b)=\frac{a^m+b^n+(a+b)^m}{a^n+b^n+(a+b)^n}\)。问在什么情况下,\(f(a,b)\)是一个整式?

比如说\(m=4\),\(n=2\)。
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2023-12-5 11:39:49 | 显示全部楼层
初筛如下:
  1. f[a_,b_,m_,n_]:=(a^m+b^m+(a+b)^m)/(a^n+b^n+(a+b)^n);
  2. x=Select[Subsets[Range@200,{2}],IntegerQ[f[2,1,#[[2]],#[[1]]]]&];
  3. y=Select[x,IntegerQ[f[3,2,#[[2]],#[[1]]]]&];
  4. z=Select[y,IntegerQ[f[7,3,#[[2]],#[[1]]]]&]
复制代码


{{2,4},{2,8},{2,10},{2,14},{2,16},{2,20},{2,22},{2,26},{2,28},{2,32},{2,34},{2,38},{2,40},{2,44},{2,46},{2,50},{2,52},{2,56},{2,58},{2,62},{2,64},{2,68},{2,70},{2,74},{2,76},{2,80},{2,82},{2,86},{2,88},{2,92},{2,94},{2,98},{2,100},{2,104},{2,106},{2,110},{2,112},{2,116},{2,118},{2,122},{2,124},{2,128},{2,130},{2,134},{2,136},{2,140},{2,142},{2,146},{2,148},{2,152},{2,154},{2,158},{2,160},{2,164},{2,166},{2,170},{2,172},{2,176},{2,178},{2,182},{2,184},{2,188},{2,190},{2,194},{2,196},{2,200},{4,10},{4,16},{4,22},{4,28},{4,34},{4,40},{4,46},{4,52},{4,58},{4,64},{4,70},{4,76},{4,82},{4,88},{4,94},{4,100},{4,106},{4,112},{4,118},{4,124},{4,130},{4,136},{4,142},{4,148},{4,154},{4,160},{4,166},{4,172},{4,178},{4,184},{4,190},{4,196}}

$(m,n)=(2k,2),k%3!=0$ or $(m,n)=(6k+4,4)$
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2023-12-5 11:43:27 | 显示全部楼层
$(m,n)={(6k+2,2),(6k+4,2),(6k+4,4)\},k >=0,k \in Z$
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2023-12-5 11:57:21 | 显示全部楼层
  1. ClearAll["Global`*"]; Table[{k,
  2.   FactorList[a^(2 + 6 k) + b^(2 + 6 k) + (a + b)^(2 + 6 k)]}, {k, 8}]
复制代码


{{1,{{2,1},{a^2+a b+b^2,1},{a^6+3 a^5 b+10 a^4 b^2+15 a^3 b^3+10 a^2 b^4+3 a b^5+b^6,1}}},{2,{{1,1},{a^2+a b+b^2,1},{2 a^12+12 a^11 b+77 a^10 b^2+275 a^9 b^3+649 a^8 b^4+1078 a^7 b^5+1276 a^6 b^6+1078 a^5 b^7+649 a^4 b^8+275 a^3 b^9+77 a^2 b^10+12 a b^11+2 b^12,1}}},{3,{{1,1},{a^2+a b+b^2,1},{2 a^18+18 a^17 b+170 a^16 b^2+952 a^15 b^3+3723 a^14 b^4+10829 a^13 b^5+24208 a^12 b^6+42483 a^11 b^7+59279 a^10 b^8+66198 a^9 b^9+59279 a^8 b^10+42483 a^7 b^11+24208 a^6 b^12+10829 a^5 b^13+3723 a^4 b^14+952 a^3 b^15+170 a^2 b^16+18 a b^17+2 b^18,1}}},{4,{{1,1},{a^2+a b+b^2,1},{2 a^24+24 a^23 b+299 a^22 b^2+2277 a^21 b^3+12374 a^20 b^4+51129 a^19 b^5+166727 a^18 b^6+439944 a^17 b^7+955604 a^16 b^8+1729002 a^15 b^9+2627129 a^14 b^10+3370029 a^13 b^11+3660542 a^12 b^12+3370029 a^11 b^13+2627129 a^10 b^14+1729002 a^9 b^15+955604 a^8 b^16+439944 a^7 b^17+166727 a^6 b^18+51129 a^5 b^19+12374 a^4 b^20+2277 a^3 b^21+299 a^2 b^22+24 a b^23+2 b^24,1}}},{5,{{2,1},{a^2+a b+b^2,1},{a^30+15 a^29 b+232 a^28 b^2+2233 a^27 b^3+15515 a^26 b^4+82940 a^25 b^5+354641 a^24 b^6+1245347 a^23 b^7+3659162 a^22 b^8+9119891 a^21 b^9+19477067 a^20 b^10+35915282 a^19 b^11+57504071 a^18 b^12+80267447 a^17 b^13+97946282 a^16 b^14+104647631 a^15 b^15+97946282 a^14 b^16+80267447 a^13 b^17+57504071 a^12 b^18+35915282 a^11 b^19+19477067 a^10 b^20+9119891 a^9 b^21+3659162 a^8 b^22+1245347 a^7 b^23+354641 a^6 b^24+82940 a^5 b^25+15515 a^4 b^26+2233 a^3 b^27+232 a^2 b^28+15 a b^29+b^30,1}}},{6,{{1,1},{a^2+a b+b^2,1},{2 a^36+36 a^35 b+665 a^34 b^2+7735 a^33 b^3+65415 a^32 b^4+428792 a^31 b^5+2266474 a^30 b^6+9924990 a^29 b^7+36712028 a^28 b^8+116374622 a^27 b^9+319647106 a^26 b^10+767300560 a^25 b^11+1620527482 a^24 b^12+3027122254 a^23 b^13+5021904364 a^22 b^14+7422259942 a^21 b^15+9795810124 a^20 b^16+11563073314 a^19 b^17+12219117172 a^18 b^18+11563073314 a^17 b^19+9795810124 a^16 b^20+7422259942 a^15 b^21+5021904364 a^14 b^22+3027122254 a^13 b^23+1620527482 a^12 b^24+767300560 a^11 b^25+319647106 a^10 b^26+116374622 a^9 b^27+36712028 a^8 b^28+9924990 a^7 b^29+2266474 a^6 b^30+428792 a^5 b^31+65415 a^4 b^32+7735 a^3 b^33+665 a^2 b^34+36 a b^35+2 b^36,1}}},{7,{{1,1},{a^2+a b+b^2,1},{2 a^42+42 a^41 b+902 a^40 b^2+12300 a^39 b^3+122549 a^38 b^4+951159 a^37 b^5+5985344 a^36 b^6+31384065 a^35 b^7+139863218 a^34 b^8+537683225 a^33 b^9+1803710335 a^32 b^10+5327945572 a^31 b^11+13959026706 a^30 b^12+32628554154 a^29 b^13+68368227668 a^28 b^14+128914835234 a^27 b^15+219431743012 a^26 b^16+338007219730 a^25 b^17+472091734222 a^24 b^18+598732526104 a^23 b^19+690215089744 a^22 b^20+723668784232 a^21 b^21+690215089744 a^20 b^22+598732526104 a^19 b^23+472091734222 a^18 b^24+338007219730 a^17 b^25+219431743012 a^16 b^26+128914835234 a^15 b^27+68368227668 a^14 b^28+32628554154 a^13 b^29+13959026706 a^12 b^30+5327945572 a^11 b^31+1803710335 a^10 b^32+537683225 a^9 b^33+139863218 a^8 b^34+31384065 a^7 b^35+5985344 a^6 b^36+951159 a^5 b^37+122549 a^4 b^38+12300 a^3 b^39+902 a^2 b^40+42 a b^41+2 b^42,1}}},{8,{{1,1},{a^2+a b+b^2,1},{2 a^48+48 a^47 b+1175 a^46 b^2+18377 a^45 b^3+210748 a^44 b^4+1889635 a^43 b^5+13790317 a^42 b^6+84204448 a^41 b^7+438883885 a^40 b^8+1982345367 a^39 b^9+7851048918 a^38 b^10+27520344515 a^37 b^11+86028257667 a^36 b^12+241311916418 a^35 b^13+610505482215 a^34 b^14+1399012176487 a^33 b^15+2914172036873 a^32 b^16+5534195177790 a^31 b^17+9605161669112 a^30 b^18+15266586536298 a^29 b^19+22257464038550 a^28 b^20+29803395487952 a^27 b^21+36688955738098 a^26 b^22+41550902139550 a^25 b^23+43308802158652 a^24 b^24+41550902139550 a^23 b^25+36688955738098 a^22 b^26+29803395487952 a^21 b^27+22257464038550 a^20 b^28+15266586536298 a^19 b^29+9605161669112 a^18 b^30+5534195177790 a^17 b^31+2914172036873 a^16 b^32+1399012176487 a^15 b^33+610505482215 a^14 b^34+241311916418 a^13 b^35+86028257667 a^12 b^36+27520344515 a^11 b^37+7851048918 a^10 b^38+1982345367 a^9 b^39+438883885 a^8 b^40+84204448 a^7 b^41+13790317 a^6 b^42+1889635 a^5 b^43+210748 a^4 b^44+18377 a^3 b^45+1175 a^2 b^46+48 a b^47+2 b^48,1}}}}

点评

nyy
你这个是穷举法吗  发表于 2023-12-5 13:06
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2023-12-5 13:10:07 | 显示全部楼层
分子有没写错,是b的m次方还是n次方
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2023-12-5 17:16:15 | 显示全部楼层
本帖最后由 northwolves 于 2023-12-5 20:54 编辑

$令a = b s 则有 (a^(2 + 6 k) + b^(2 + 6 k) + (a + b)^(2 + 6 k))/(a^2 +     b^2 + (a + b)^2)$

$=\frac{b^{6 k+2}+(b s)^{6 k+2}+(b (s+1))^{6 k+2}}{2 b^2 \left(s^2+s+1\right)}$

$=b^(6k)/2*\frac{s^{6 k+2}+(s+1)^{6 k+2}+1}{s^2+s+1}$

$=b^(6k)/2*\frac{(s-1)(s^{6 k+2}+(s+1)^{6 k+2}+1)}{(s-1)(s^2+s+1)}$


$=b^(6k)/2*\frac{(s^{3 + 6 k} -1)-s^2(s^(6 k) -1)+(s -1) (s + 1)^2 ((s + 1)^(6 k) - 1) + (s^3 - 1)}{s^3-1}$

$=b^(6k)/2*(1+\frac{((s^3)^{2k+1} -1)-s^2((s^3)^(2 k) -1)+(s -1) (s + 1)^2 (((s + 1)^6)^k - 1) }{s^3-1})$

点评

$\frac{(s-1) \left((s+1)^6-1\right)}{s^3-1}=s^4+5 s^3+9 s^2+6 s$  发表于 2023-12-5 17:43
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2023-12-5 20:53:21 | 显示全部楼层
$\frac{a^4+(a+b)^4+b^4}{a^2+(a+b)^2+b^2}=a^2+a b+b^2$

$\frac{a^{10}+(a+b)^{10}+b^{10}}{a^4+(a+b)^4+b^4}=a^6 + 3 a^5 b + 27/2 a^4 b^2 + 22 a^3 b^3 + 27/2 a^2 b^4 +
3 a b^5 + b^6$
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2023-12-5 21:24:24 | 显示全部楼层
$令a = b s 则有 (a^(4 + 6 k) + b^(4 + 6 k) + (a + b)^(4 + 6 k))/(a^4 + b^4 + (a + b)^4)$

$=\frac{b^{6 k+4}+(b s)^{6 k+4}+(b (s+1))^{6 k+4}}{2 b^4 \left(s^2+s+1\right)^2}$

$=b^(6k)/2*\frac{s^{6 k+4}+(s+1)^{6 k+4}+1}{(s^2+s+1)^2}$

点评

这里不太好分解,可以数学归纳法证明之。  发表于 2023-12-5 21:26
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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