答案是:不.
这里是最小的反例:取$n=29\cdot7645373=221715817=29\cdot197^3$。
然后$\phi(n)=212983792$和$28\cdot\phi。
我用以下方法找到了这个平价单线:
? 对于(i=010000000,如果(!istotient(28*eulerphi(29*i)),打印(i))7645373
编辑:再往下看,有很多$n=29\cdot197^3\cdoti形式的反例$
? 对于(i=1,+oo,如果(!istotient(28*eulerphi(29*197^3*i)),打印1(i);打印1(“,”))1,2,3,4,6,7517,15034,18059,22551,28019,30068,30983,36118,45102,56038,61966,65267,67427,67499,71387,84057,84947,90677,92949,97187,112076,115469,123932,127487,130534,130787,134854,134998,142774,168114,169067,169894,181354,185898,191579,194374,195801,202281,214161,218627,227519,230189,230938,233939,239843,247847,252029,253367,254841,254974,261068,261574,262643,269708,272031,285548,289937,291561,294059,297707,330587,338134,339788,339827,347003,349343,350159,362708,381917,383158,387503,388748,391602,392361,404562,405683,427247,428322,437254,440039,447179,454907,455038,460378,467783,467878,475367,479686,489197,495694,504058,506734,507201,509682,514247,523148,525286,544062,546719,548837,548903,564797,569813,574737,579287,579874,583122,588118,595414,597383,608117,628379,635939,644597,647579,654527,655881,661174,664019,676268,679654,682557,690567,694006,697259,698686,700318,701817,714827,718139,723287,739163,743541,748019,750719,760101,763013,763834,766316,775006,776267,781307,784722,787929,800399,804179,811366,817319,829847,832787,847163,850727,854494,863887,874508,877403,878987,880078,880799,882083,889589,894358,901997,909814,910076,920756,935566,935756,950734,962747,974837,978203,978394,979919,980327,981287,989123,991388,991761,1013468,1014402,1017923,1019481,1028494,1030307,1036307,1041009,1048029,1050572,1053293,1053407,1070039,1082717,1093438,1097674,1097806,1119299,1129594,1138637,1139626
如果您正在寻找更大的$n$,可以采用形式为$n=29\cdot 197^14\cdot i$的数字来表示$i\in\{1,2116387232774,…\}$
这些小案例让人觉得似乎有道理,因为有无数反例。你能找到一个无限的家庭吗?有没有反例,不能除以197^3$?