The question is to judge whether $4^{351012345} \equiv 5^{543213510}$ mod $231$ is true. I'm thinking of just directly calculating the exact modulus of both numbers using modular exponentiation but is there a better way to prove the equivalence holds using another method?
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1Do you know Euler's theorem? – J. W. Tanner Sep 13 '20 at 20:04
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1Using Euler's theorem is the standard way to proceed once you know that $\varphi(231)=\varphi(3\cdot7\cdot11)=120$. – Andrea Mori Sep 13 '20 at 20:04
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I didn't get very far with it, but there were some interesting results by transforming the exponents from base $10$ into the base of the modulus. This has a very similar effect to modular exponentiation of course. – abiessu Sep 13 '20 at 20:05
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3Since $231= 3\cdot 7 \cdot 11$, all you need to do is prove the two numbers are congruent modulo each of the primes. That makes it pretty quick. – B. Goddard Sep 13 '20 at 20:06
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How do you define as "a better way?" Start with direct calculation, and show us that work. You need to provide more context. – amWhy Sep 13 '20 at 20:06
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As B. Goddard comments, the quick approach it to work separately modulo each of the prime factors (with the help of Fermat's little theorem). Equivalently, we could use the Chinese remainder theorem – Ben Grossmann Sep 13 '20 at 20:08
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The Carmichael function of $231$ is lcm$(3-1,7-1,11-1)=30$. – J. W. Tanner Sep 13 '20 at 20:10
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As alluded to in comments, the prime factorization of $231=3\times7\times11$.
The Carmichael function of $231$ is thus lcm$(3-1,7-1,11-1)=30$.
Therefore, we merely have to prove that $4^{15}\equiv5^0=1\bmod231$,
which follows from $4\equiv1\bmod3, 4^3\equiv1\bmod7$, and $4^5\equiv1\bmod11$.
J. W. Tanner
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Unlikely the OP already knows about Carmichael since it is not often presented (early) in elementary number theory textbooks – Bill Dubuque Sep 13 '20 at 20:16
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Hint $\bmod m=3,7, 11\!:\ 4^{\large 15n}\! = 2^{\large \color{#c00}{30}n}\!\equiv 1\equiv 5^{\large \color{#c00}{30}k}\,$ by $\,\phi(m)\mid \color{#c00}{30}\,$ & mod order reduction. Now apply CRT (or CCRT, $ $ i.e. $\,4^{\large 15n}\!- 5^{\large {30}k^{\phantom{|}}}\!$ is divisible by $3,7,11$ so also by their lcm = product = $231$).
Bill Dubuque
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1Note: by casting out 3's both big exponents are divisible by $3$ since their digit sum is, and via last digits the expt on $4$ is divisible by $5$, and the expt on $5$ by $10$, so the expt on $4$ is divisible by $15$ and the expt on $5$ by $30$, all by trivial mental arithmetic. – Bill Dubuque Sep 13 '20 at 20:42