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Fermat's Little Theorem states that whenever \( n \) is prime and \( a \) is an integer, \[ a^{n-1} \equiv 1 \pmod{n} \]

Given \( a = 41 \) and \( n = 139 \), efficiently compute \( 41^{138} \equiv 1 \pmod{139} \).

A) Calculate modulo 139.
B) Determine modulo 139 using the theorem.
C) Verify Fermat's Little Theorem with \( a = 41 \) and \( n = 139 \).

Answer :

Final Answer:

Fermat's Little Theorem states that for any prime number n and integer \( a ), [tex]\( a^{n-1} \equiv 1 \pmod{n} \)[/tex]. In this case, with a = 4 and n = 139, we efficiently compute[tex]\( 4^{138} \pmod{139} \)[/tex], resulting in the congruence [tex]\( 4^{138} \equiv 1 \pmod{139} \)[/tex].

Explanation:

Now, let's delve into the explanation. Fermat's Little Theorem asserts that if n is a prime number, then [tex]\( a^{n-1} \equiv 1 \pmod{n} \)[/tex] for any integer a not divisible by n. In this scenario,( a = 4 ) and ( n = 139 ).

Therefore, [tex]\( 4^{138} \equiv 1 \pmod{139} \)[/tex] by Fermat's Little Theorem. To further elaborate, (4¹³⁸) is congruent to 1 modulo 139, indicating that when 139 divides (4¹³⁸ - 1 \), there is no remainder.

This can be expressed as [tex]\( 4^{138} \equiv 1 \pmod{139} \)[/tex]. The application of Fermat's Little Theorem simplifies the calculation of the congruence.

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