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Peter
Birch & Swinnerton Dyer
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Post Posted: May 24, 2007, 5:24 pm • # 1 


Prove that for n\geq 2, \underbrace{2^{2^{\cdots^{2}}}}_{n\text{ terms}}\equiv \underbrace{2^{2^{\cdots^{2}}}}_{n-1\text{ terms}}\; \pmod{n}.
 
 
ZetaX
Birch & Swinnerton Dyer
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Post Posted: May 24, 2007, 5:24 pm • # 2 


Taking a closer look at my proof in http://www.artofproblemsolving.com/viewtopic.php?p=849418.
 
 
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Community » Olympiad Section » Number Theory » PEN » Official PEN problems » Arithmetic in Zn » Congruences

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