Do modular rules apply for repeated exponents

Asked Apr 01 '20 at 15:45

Active Apr 01 '20 at 15:45

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The exponent rule for modular arithmetic states that $$a^b \equiv (a \;mod \; m)^b \;(mod\;m)$$

Does this rule apply for repeated exponents which are computed from top to bottom? For example, is this true? $$a^{b^c} \equiv (a\;mod\;m)^{(b\;mod\;m)^c}\;(mod\;m)$$

asked Apr 01 '20 at 15:45

pblpbl

Euler's theorem says exponents work mod $\varphi(n)$ when you work mod $n$ in the base. – Randall Apr 01 '20 at 15:47
No, in exponents you need to apply modular order reduction – Bill Dubuque Apr 01 '20 at 15:55
e.g. $\large \bmod p!:\ a \equiv a^p \not\equiv a^0,$ generally. – Bill Dubuque Apr 01 '20 at 15:59
Due to the first statement and because exponentiation starts from top $$b^c \equiv (b ;mod ; m)^c ;(mod;m)$$ therefore the second statement should be $$a^{b^c} \equiv (a;mod;m)^{(b ;mod ; m)^c ;(mod;m)};(mod;m)$$ I guess. – Dávid Laczkó Apr 01 '20 at 16:01
See here for one of many prior answers on power tower modular reductions. – Bill Dubuque Apr 01 '20 at 16:04
@DávidLaczkó That is not correct. – Bill Dubuque Apr 01 '20 at 16:04

Do modular rules apply for repeated exponents

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