Collection of Maths Problems

Powers in finite fields

Task number: 2482

• Variant 1

  Determine values \(2^{101},3^{1\,001}\) and \(4^{1\,000\,001}\) in the field \(\mathbb Z_{17}\).

• Hint

  Calculate powers of the base and observe when you get the unit.

• Resolution

  \(2^8=1\), hence \(2^{101}=2^{8{\cdot} 12+5}=(2^8)^{12}\cdot 2^5=1^{12}\cdot 15 = 15\).

  \(3^{16}=1\) hence \(3^{1001}=3^{16{\cdot} 62+9}=(3^{16})^{62}\cdot 3^9=1^{62}\cdot 14 = 14\).

  \(4^{4}=1\) hence \(4^{1000001}=4^{4{\cdot} 250000+1}=(4^4)^{250000}\cdot 4^1=1^{250000}\cdot 4 = 4\).
  

• Variant 2

  Determine \(5^{100},8^{200},11^{300}\) and \(18^{400}\) in the field \(\mathbb Z_{19}\).

• Resolution

  \(5^9=1\) hence \(5^{100}=1{\cdot} 5^1 = 5\).

  \(8^6=1\) hence \(8^{200}=1{\cdot} 8^2 = 7\).

  \(11^3=1\) hence \(11^{300}=(11^3)^{100} = 1\).

  \(18^2=1\) hence \(18^{400}=(18^2)^{200} = 1\). \medskip

  By the way, by the Fermat's little theorem \[\forall {\rm prime}\ p, \forall a\in \mathbb N: \quad p\nmid a \quad\Longrightarrow\quad a^{p-1}\equiv 1 \quad({\rm mod\ } p)\] follows, that exponents in \(\mathbb Z_p\) could always be reduced modulo \(p-1\).