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Old 2003-07-30, 14:52   #1
hyh1048576
 
Jun 2003

26 Posts
Default Hey!Have a try?

Surpose a is the largest root of equation x^3-3x^2+1=0
Show that [a^2003]=6(mod17)(6.17 is my mother's birthday :D ), here [] is the floor function.

It won't be too difficult :)
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Old 2003-08-20, 10:27   #2
wpolly
 
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Sep 2002
Vienna, Austria

21910 Posts
Default

Here is my solution(Brute-Force-Calculate):
[code:1]
In[1]:= a = 2 Cos[Pi/9]+1;
In[2]:= a^3-3*a^2+1//FullSimplify
Out[2]:= 0
In[3]:= Floor[a^2003]
Out[3]:=
94939725522917145069342477120927832559217730051429875325875570636924182335675242126888987465659418824501369141352027924590623201155390783435192180777593185869652542874753609165228507858331675127044297709318074826862717121658557051997230571700012789454391723891537541722772990652755472059116333668130669428008299627650443250600118347941504774401511914331708152817467437650285026441264284293647908581851475235662683990931712031371117366408062644550668223083770926729445479986852972153706088341046916910492234320780615522305461555252832852742243918512824820924363220806936531149542424166231769975014115743842273168608484003530964515830286198897621954981641073966910325867573555031727328537616848211999480772476042265593242575128038332335279596926387211744794119883009475921654997485217025559320699124433598243301894200698469075388466837907563839880516118870350655626346555112096882637672985509380270107651106836649660622590
In[4]:= Mod[%, 17]
Out[4]:= 6
[/code:1]

Q. E. D.
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