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x^(n)+y^(n+1)+z^(n+2)=w^(n+3)

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发表于 2023-2-22 22:28 | 显示全部楼层
本帖最后由 Treenewbee 于 2023-2-22 22:43 编辑

\[n=2k, \frac {((t^{n + 3} - 2)^{\frac{(n + 2) (n + 3)}{2}})^{n} + ((t^{n + 3}-2)^{\frac{(n^2 + 4 n + 2)}{2}})^{n + 1}+ ((t^{n + 3}-2)^{\frac{n (n + 3)}{2}})^{n + 2} } {(t (t^{n + 3} - 2)^{\frac{n (n + 2)}{2}})^{n + 3}}=1\]

\[n=2k+1, \frac {((t^{n+3}-2)^{\frac{(n+1)^2(n+3)}{2}})^n+((t^{n+3}-2)^{\frac{n(n+1)(n+3)}{2}})^{n+1}+((t^{n+3}-2)^{\frac{(n^3+3n^2+n+1)}{2}})^{n+2}}{(t(t^{n+3}-2)^{\frac{n(n+1)^2}{2}})^{n+3}}=1\]

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参与人数 1威望 +20 收起 理由
王守恩 + 20 很给力!看懂了!

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 楼主| 发表于 2023-2-24 11:14 | 显示全部楼层
加数指数是连续 5 个的,还有吗?会有吗?!

\((4^{020})^{1}+(4^{010})^{2}+(4^{005})^{04}+(4^{004})^{05}=(4^{07})^{03}\)
\((4^{045})^{2}+(4^{030})^{3}+(4^{018})^{05}+(4^{015})^{06}=(4^{13})^{07}\)
\((4^{084})^{3}+(4^{063})^{4}+(4^{042})^{06}+(4^{036})^{07}=(4^{23})^{11}\)
\((4^{140})^{4}+(4^{112})^{5}+(4^{080})^{07}+(4^{070})^{08}=(4^{33})^{17}\)
\((4^{216})^{5}+(4^{180})^{6}+(4^{135})^{08}+(4^{120})^{09}=(4^{47})^{23}\)
\((4^{315})^{6}+(4^{270})^{7}+(4^{210})^{09}+(4^{189})^{10}=(4^{61})^{31}\)
\((4^{440})^{7}+(4^{385})^{8}+(4^{308})^{10}+(4^{280})^{11}=(4^{79})^{39}\)
\((4^{594})^{8}+(4^{528})^{9}+(4^{432})^{11}+(4^{396})^{12}=(4^{97})^{49}\)
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发表于 2023-2-24 15:39 | 显示全部楼层
\[(4^{24})^1+(4^{12})^2+(4^{8})^3+(4^{6})^4=(4^{5})^5\]

\[(4^{240})^3+(4^{180})^4+(4^{144})^5+(4^{120})^6=(4^{103})^7\]

\[(4^{3600})^7+(4^{3150})^8+(4^{2800})^9+(4^{2520})^{10}=(4^{2291})^{11}\]

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傻了!把=改+,再添一个=,不就是:5个加数,指数连续,!谢谢!!!  发表于 2023-2-24 17:41

评分

参与人数 1威望 +20 收起 理由
王守恩 + 20 5个加数,指数连续, 会有吗?!

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发表于 2023-2-24 21:49 | 显示全部楼层
\[(5^{12240})^{14}+(5^{11424})^{15}+(5^{10710})^{16}+(5^{10080})^{17}+(5^{9520})^{18}=(5^{9019})^{19}\]
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发表于 2023-2-24 21:50 | 显示全部楼层
\[(5^{122850})^{24}+(5^{117936})^{25}+(5^{113400})^{26}+(5^{109200})^{27}+(5^{105300})^{28}=(5^{101669})^{29}\]
\[(5^{1945800})^{44}+(5^{1902560})^{45}+(5^{1861200})^{46}+(5^{1821600})^{47}+(5^{1783650})^{48}=(5^{1747249})^{49}\]
\[(5^{5091240})^{54}+(5^{4998672})^{55}+(5^{4909410})^{56}+(5^{4823280})^{57}+(5^{4740120})^{58}=(5^{4659779})^{59}\]
\[(5^{22822800})^{74}+(5^{22518496})^{75}+(5^{22222200})^{76}+(5^{21933600})^{77}+(5^{21652400})^{78}=(5^{21378319})^{79}\]
\[(5^{41974020})^{84}+(5^{41480208})^{85}+(5^{40997880})^{86}+(5^{40526640})^{87}+(5^{40066110})^{88}=(5^{39615929})^{89}\]
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发表于 2023-2-24 21:50 | 显示全部楼层
\[(5^{117900090})^{104}+(5^{116777232})^{105}+(5^{115675560})^{106}+(5^{114594480})^{107}+(5^{113533420})^{108}=(5^{112491829})^{109}\]
\[(5^{184172040})^{114}+(5^{182570544})^{115}+(5^{180996660})^{116}+(5^{179449680})^{117}+(5^{177928920})^{118}=(5^{176433719})^{119}\]
\[(5^{404966520})^{134}+(5^{401966768})^{135}+(5^{399011130})^{136}+(5^{396098640})^{137}+(5^{393228360})^{138}=(5^{390399379})^{139}\]
\[(5^{575718150})^{144}+(5^{571747680})^{145}+(5^{567831600})^{146}+(5^{563968800})^{147}+(5^{560158200})^{148}=(5^{556398749})^{149}\]
\[(5^{1088642940})^{164}+(5^{1082045104})^{165}+(5^{1075526760})^{166}+(5^{1069086480})^{167}+(5^{1062722870})^{168}=(5^{1056434569})^{169}\]
\[(5^{1455577200})^{174}+(5^{1447259616})^{175}+(5^{1439036550})^{176}+(5^{1430906400})^{177}+(5^{1422867600})^{178}=(5^{1414918619})^{179}\]
\[(5^{2484682200})^{194}+(5^{2471940240})^{195}+(5^{2459328300})^{196}+(5^{2446844400})^{197}+(5^{2434486600})^{198}=(5^{2422252999})^{199}\]
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发表于 2023-2-24 22:07 | 显示全部楼层
\[(2^{168})^{6}+(2^{144})^{7}+(2^{126})^{8}+(2^{112})^{9}=(2^{101})^{10}\]
\[(2^{858})^{10}+(2^{780})^{11}+(2^{715})^{12}+(2^{660})^{13}=(2^{613})^{14}\]
\[(2^{6650})^{18}+(2^{6300})^{19}+(2^{5985})^{20}+(2^{5700})^{21}=(2^{5441})^{22}\]
\[(2^{13800})^{22}+(2^{13200})^{23}+(2^{12650})^{24}+(2^{12144})^{25}=(2^{11677})^{26}\]
\[(2^{43648})^{30}+(2^{42240})^{31}+(2^{40920})^{32}+(2^{39680})^{33}=(2^{38513})^{34}\]
\[(2^{69930})^{34}+(2^{67932})^{35}+(2^{66045})^{36}+(2^{64260})^{37}=(2^{62569})^{38}\]
\[(2^{156090})^{42}+(2^{152460})^{43}+(2^{148995})^{44}+(2^{145684})^{45}=(2^{142517})^{46}\]
\[(2^{221088})^{46}+(2^{216384})^{47}+(2^{211876})^{48}+(2^{207552})^{49}=(2^{203401})^{50}\]
\[(2^{409640})^{54}+(2^{402192})^{55}+(2^{395010})^{56}+(2^{388080})^{57}=(2^{381389})^{58}\]
\[(2^{539850})^{58}+(2^{530700})^{59}+(2^{521855})^{60}+(2^{513300})^{61}=(2^{505021})^{62}\]
\[(2^{890698})^{66}+(2^{877404})^{67}+(2^{864501})^{68}+(2^{851972})^{69}=(2^{839801})^{70}\]
\[(2^{1119528})^{70}+(2^{1103760})^{71}+(2^{1088430})^{72}+(2^{1073520})^{73}=(2^{1059013})^{74}\]
\[(2^{1706400})^{78}+(2^{1684800})^{79}+(2^{1663740})^{80}+(2^{1643200})^{81}=(2^{1623161})^{82}\]
\[(2^{2074170})^{82}+(2^{2049180})^{83}+(2^{2024785})^{84}+(2^{2000964})^{85}=(2^{1977697})^{86}\]
\[(2^{2984618})^{90}+(2^{2951820})^{91}+(2^{2919735})^{92}+(2^{2888340})^{93}=(2^{2857613})^{94}\]
\[(2^{3538560})^{94}+(2^{3501312})^{95}+(2^{3464840})^{96}+(2^{3429120})^{97}=(2^{3394129})^{98}\]

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参与人数 1威望 +10 收起 理由
王守恩 + 10 哇塞!吓坏宝宝了!

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 楼主| 发表于 2023-2-25 08:35 | 显示全部楼层
谢谢 Treenewbee!

任意(a)个加数,且指数(n)连续, 都可以有的。

  先立座灯塔,航行大海就有方向了。

\(\displaystyle\frac{1}{a^{\frac{n!+(n+a)!}{n!}}}\sum_{k=n+1}^{n+a}\bigg(a^{\frac{(n+a)!}{n!k}}\bigg)^{k}=1\)


a=1, 2, 3, 4, 5, 6, 7, 8, 9, ......       n=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
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 楼主| 发表于 2023-2-25 09:16 | 显示全部楼层
\((4^{020})^{1}+(4^{010})^{2}+(4^{005})^{04}+(4^{004})^{05}=(4^{07})^{03}\)
\((4^{045})^{2}+(4^{030})^{3}+(4^{018})^{05}+(4^{015})^{06}=(4^{13})^{07}\)
\((4^{084})^{3}+(4^{063})^{4}+(4^{042})^{06}+(4^{036})^{07}=(4^{23})^{11}\)
\((4^{140})^{4}+(4^{112})^{5}+(4^{080})^{07}+(4^{070})^{08}=(4^{33})^{17}\)
\((4^{216})^{5}+(4^{180})^{6}+(4^{135})^{08}+(4^{120})^{09}=(4^{47})^{23}\)
\((4^{315})^{6}+(4^{270})^{7}+(4^{210})^{09}+(4^{189})^{10}=(4^{61})^{31}\)
\((4^{440})^{7}+(4^{385})^{8}+(4^{308})^{10}+(4^{280})^{11}=(4^{79})^{39}\)
\((4^{594})^{8}+(4^{528})^{9}+(4^{432})^{11}+(4^{396})^{12}=(4^{97})^{49}\)

\(\frac{\big(4^{(n^2-1)(n+2)/2}\big)^{n-2}+\big(4^{(n^2-4)(n+1)/2}\big)^{n-1}+\big(4^{(n^2-4)(n-1)/2}\big)^{n+1}+\big(4^{(n^2-1)(n-2)/2}\big)^{n+2}}{\big(4^{(2n^2-5-\cos(n\pi))/2}\big)^{(2n^2-5+\cos(n\pi))/4}}=1\)
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 楼主| 发表于 2023-3-5 08:44 | 显示全部楼层
本帖最后由 王守恩 于 2023-3-5 08:48 编辑
Treenewbee 发表于 2023-2-7 18:27
\[n=2, 16^2+8^3+4^4=4^5\]

尊敬的Treenewbee!请教一个问题。

\(题目:若x,y,d为正整数,\ \ x^2-dy^2=1, \ \ 则d=1, 4, 9, 16, 25, 36,.....无解(答案有问题)。\)

Select[Union@Flatten@Table[(y^2-1)/x^2,{x,1,50},{y,2,Sqrt[50x^2+1]}],IntegerQ[#]&]

{2, 3, 5, 6, 7, 8, 10, 11, 12, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26,
27, 28, 30, 32, 33, 34, 35, 37, 38, 39, 40, 42, 44, 45, 47, 48, 50}

问:x 如何取值?

点评

部分d未取到是因为x取值不夠大,比如d=13,{x->649,y->180};d=61,{x->1766319049,y->226153980}  发表于 2023-3-5 09:22
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