Japanese mathematician Sin Hitotsumatsu asked the proof or the contradiction that, apart from tribial solutions 102n, 4*102n and 9*102n, there are perfect square number consisted of only 2 different decimal digits. Known largest solution is 816192 = 6661661161.

I extended the condition of this problem as (i) 3 different digits (ii) perfect n-th power, and found many remarkable parametrized patterns and sporadic solutions.

All solutions of sporadic pattern up to 1025 for non-zero patterns

and up to 1023 patterns including zero are here.

1501674067666649999852= 22550250055025025225250200022000050000225 (April 07, 2008)

1490670655108730886732= 22220990020022929092929022220290920900929 (April 07, 2008)



31802522547770395385022= 10114004404014444004140001011411401140404004 (April 28, 2008)

66749834797132300059622= 44555404454444540454555540045000554555545444 (June 02, 2008)

30157752651590112301382= 9094900449944904494440090444449999999499044 (June 09, 2008)



449499949999999499999952= 2020502050500020505000050500052500000500000025 (October 31, 2008)

208327397238179751383622= 434003044400343443044430000434430333044043044 (October 31, 2008)

All solutions of infinite pattern (parameterized solutions) are here.

Sometimes the first (smallest) solution becomes very large. For example, 056 : 22360814084166662 = 5000060065066660656065066555556 (May 4, 1997)

079 : 88191722853734972 = 77777799799099990007000790009009 (May 5, 1997)

789 : 99493707779879172 = 98989978877879888789778997998889 (May 10, 1997)

019 : 436942788245669642512 = 1909190001999001011109190090109911991001 (May 6, 1998)

I searched up to the following range, but couldn't find the solution of 013 and 689; 013 : 1024

678 : 1025