31 lines
691 B
Markdown
31 lines
691 B
Markdown
#Math #NT
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# Theorem
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Let $a$ and $m$ be coprime numbers.
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$$
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a^{\phi(m)} \equiv 1 \mod m
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$$
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This is a generalization of Fermet's Little Theorem, as $m$ is a prime number in Fermet’s Little Theorem.
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# Proof
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Let:
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$$
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A = \{p_1, p_2, p_3,... p_{\phi(m)} \} \mod m \\
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B = \{ap_1, ap_2, ap_3,...ap_{\phi(m)}\} \mod m
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$$
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Where $p_x$ is the $x$th number relatively prime to $m$.
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Since $a$ and $p_x$ are coprime to $m$, $ap_x$ is coprime to $m$. Since each $p_x$ is unique, $ap_x$ is unique, which makes set $B$ the same as set $A$.
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Since all terms are coprime to $m$:
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$$
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a^{\phi(m)} \prod _{k = 0}^{\phi(m)} p_k \equiv \prod _{k = 0}^{\phi(m)} p_k \mod m \\
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a^{\phi(m)} \equiv 1 \mod m
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$$ |