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# [讨论] 一个完全由 0 和 1 组成的数是否有可能为完全平方数？

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 发现其他进制下可以有很多，如： Select[Range@1000000, Max@IntegerDigits[#^2, 8] == 1 &]复制代码 八进制： {1,3,8,24,64,192,512,1536,4096,11677,12288,32768,93416,98304,262144,747328,786432} Select[Range@1000000, Max@IntegerDigits[#^2, 5] == 1 &]复制代码 五进制： {1,5,25,125,625,3125,15625,78125,390625,972799} Select[Range@1000000, Max@IntegerDigits[#^2, 7] == 1 &]复制代码 七进制： {1,7,20,49,140,343,980,2401,6860,16807,48020,117649,336140,823543} Select[Range@1000000, Max@IntegerDigits[#^2, 3] == 1 &]复制代码 三进制： {1,2,3,6,9,11,16,18,19,27,29,33,48,54,55,57,81,83,87,99,143,144,162,163,165,171,243,245,249,261,262,297,421,429,432,451,486,487,489,495,513,729,731,735,747,783,786,889,891,1263,1287,1296,1331,1342,1353,1458,1459,1461,1467,1485,1487,1539,2187,2189,2193,2205,2241,2242,2323,2349,2358,2537,2573,2644,2662,2667,2673,3788,3789,3861,3888,3993,4026,4059,4374,4375,4377,4383,4401,4455,4461,4562,4617,6561,6563,6567,6579,6615,6688,6723,6726,6967,6969,7036,7047,7074,7082,7611,7696,7719,7932,7986,8001,8019,8035,11364,11367,11583,11645,11664,11675,11979,11987,12078,12177,12202,13122,13123,13125,13131,13149,13203,13205,13310,13365,13383,13660,13678,13686,13835,13851,19683,19685,19689,19701,19737,19845,19846,20064,20089,20169,20178,20762,20791,20869,20901,20907,21108,21130,21141,21222,21246,22833,23088,23157,23671,23796,23803,23958,23974,24003,24057,24073,24105,34092,34101,34127,34749,34935,34992,35003,35025,35937,35961,36234,36531,36606,39366,39367,39369,39375,39393,39447,39609,39615,39930,40095,40149,40980,41032,41034,41058,41505,41553,41669,59049,59051,59055,59067,59103,59211,59535,59538,60192,60265,60267,60496,60507,60534,60667,62286,62373,62607,62703,62721,63324,63390,63423,63666,63738,68499,69264,69471,69550,70972,71013,71388,71409,71874,71922,71983,72009,72171,72187,72219,72315,102276,102303,102381,104164,104247,104438,104805,104976,104987,105009,105075,107811,107883,108008,108413,108494,108702,109593,109818,110374,110402,118098,118099,118101,118107,118125,118179,118341,118343,118667,118827,118845,119747,119790,120285,120447,120530,122940,122942,123096,123102,123174,124515,124659,125007,177147,177149,177153,177165,177201,177309,177633,177634,178363,178605,178614,178778,180576,180795,180801,181488,181510,181521,181602,182001,186858,187119,187804,187821,188109,188163,188260,189818,189972,190170,190269,190324,190838,190998,191176,191214,204587,205497,205928,207792,208413,208650,208753,212916,213039,213325,214148,214164,214227,215622,215766,215927,215949,216027,216513,216529,216561,216657,216945,306828,306909,307143,312492,312741,313147,313314,314415,314928,314939,314961,315027,315225,323433,323649,324024,325239,325482,325513,326106,328779,329454,330623,331122,331206,354294,354295,354297,354303,354321,354375,354537,355023,355029,356001,356481,356535,359241,359370,360855,360884,360982,361043,361341,361590,368820,368826,369288,369306,369522,370883,373545,373715,373977,375021,375707,531441,531443,531447,531459,531495,531603,531927,532899,532902,533072,535087,535089,535815,535842,536285,536334,541676,541728,542385,542403,542782,544464,544530,544563,544806,545605,546003,560574,561357,561625,563412,563463,564327,564362,564489,564780,569454,569916,570510,570807,570862,570972,572514,572994,573528,573642,613761,616491,617784,623376,623483,625239,625950,626259,638748,639117,639656,639739,639975,641843,642444,642492,642681,642790,646866,646895,647297,647298,647781,647847,648081,649539,649555,649587,649683,649971,650835,920484,920727,921429,928747,937476,938223,939441,939942,943013,943166,943245,944784,944795,944817,944883,945081,945673,945675,970299,970947,972072,975717,976446,976539,977644,978318,986337,988362,991826,991869,993366,993618,993950,994033}

### 评分

uk702 + 2 + 2 + 2 + 2 + 2 很给力!

楼主| 发表于 2024-7-10 08:20:34 | 显示全部楼层
 本帖最后由 uk702 于 2024-7-10 08:52 编辑 关于问题3，我将它发到 https://math.stackexchange.com/ （英文太烂了就不给链接了），某老师给出了如下的证法，尽管我接受了但总感觉还差点意思： 证：若 n=p，显然 $10^m + 10^n + 10^p + 1$ 是可以为完全平方数，下面讨论仅限于  m>n>p>0 的情况。
1) 如果 p = 1，则显然不会有任何完全平方数以 11 结尾，故这时 $10^m + 10^n + 10^p + 1$  不能为完全平方数。 2) 现在假设 p >= 2，若 $y^2 = 10^m + 10^n + 10^p + 1$ 为完全平方数，则 y 的个位数必须是以 1 或 9 。 a. 若 $y=10x+ 1$，则 $y^2 = 100x^2 + 20x + 1 = 10^m + 10^n + 10^p + 1$，所以 x 必须是 5 的倍数，记 x=5a。 ∴ 故必须有 $2500 \ a^2 + 100a + 1 = 10^m + 10^n + 10^p + 1$ ∴ 故必须有 $25 \ a^2 + a = 10^u + 10^v + 10^w$，其中 u>v>w ≧ 0。 【注：老师说这一行就无法成立，下面是我的补充，试图说明为什么无法成立】 记 $a = 10^k \ b$，其中 b 的个位数不为 0，k >=0，代入，则有 $(25 \ 10^{2k}) \ b^2 + 10^k \ b = 10^u + 10^v + 10^w$ ∴ w ≧ k，故有 $(25 \ 10^{k}) \ b^2 + b = 10^r + 10^s + 10^t$，其中 r>s>t>=0 。 对上式取 mod 10，∴ 必须有 b =1 (mod 10) 且 t = 0  【注，实际上，如果 k=0，b 还可以 b=5 (mod 10) 】 ∴  $(25 \ 10^{k}) \ b^2 + b = 10^r + 10^s + 1$ 由于 b 以数字 1 结尾，故左边将包含数字 5 或 6 而右边不能，故上式无法成立。 【注：我不确定这个说法一定成立】 b. 若 $y=10x - 1$，则 $y^2 = 100x^2 - 20x + 1 = 10^m + 10^n + 10^p + 1$，可参考 a) 。

楼主| 发表于 2024-7-9 20:32:47 | 显示全部楼层
 northwolves 发表于 2024-7-9 18:18 发现其他进制下可以有很多，如： 转自 http://kuing.infinityfreeapp.com ... amp;page=1#pid60547 第 5 楼。 5 进制的情形： 注意到 $(x(x(5x-1)(5x+4)+3)^2+1)^2=$ $$625 x^8 + (625+125) x^7 + 25 x^6 + (25+5) x^5 + (125+25+5+1) x^4 + (5+1) x^3 + x^2 + (5+1) x + 1.$$ 从而对较大的 $x=5^k$,  该完全平方数将只含 $0$ 与 $1$ 的平方式.

 一个n位数只有1和0的概率是$\frac{1}{5^n}$,平方数的几率$\frac{1}{\sqrt{10}^n}$。假设两个事件独立，两个条件对同一个数成立的概率为$\frac{1}{(5\sqrt{10})^n}<\frac{1}{15^n}<\frac{1}{10^n}$。

 仅由两个非0不同数字组成的平方数： {16, 25, 36, 49, 64, 81, 121, 144, 225, 441, 484, 676, 1444, 7744, 11881, 29929, 44944, 55225, 69696, 9696996, 6661661161}

### 点评

No other terms below 10^41. The sequence is probably finite.  发表于 2024-7-9 14:27

 我以前搜索到10^44，没有新值

楼主| 发表于 2024-7-9 14:54:09 | 显示全部楼层
 northwolves 发表于 2024-7-9 14:31 Sporadic solutions 三个数字的参考mathe版主大作 那么，这个问题（一个完全由 0 和两个以的 1 组成的数是否有可能为完全平方数？）是不是已经有无解的答案了？还是说尚未解决？

 无解。应该不太难证明，楼主试试看

 若$N=1******01$是一个平方数,则$\sqrtN=[13]......[19]$，N里面1的数量$n=9k+{0,1,4,7}$

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