Question

show that 16^99 - 1 = 0 ( mod 437 ) ​

Answer 1

The proof of 1$6^{99}$ - 1  = 0 (mod 437) is given below:

• To solve this problem, we will be using the following concepts:

         1) a = b (mod n) ⇒ $a^{x}$ = $b^{x}$ (mod n)

         2) If a = b (mod p) and a = b (mod q) where GCD (p , q) = 1, then

             a = b(mod pq)

• We know that 437=19×23 and GCD (19 , 23) = 1

        Now,

        (−3)³ = −8(mod 19)

        ⇒ $(-3)^{9}$ = 1 (mod 19)

        ⇒ 1$6^{9}$ = 1 (mod 19)

        ⇒ 1$6^{99}$ = 1 (mod 19)      ....................(I)

• Also,

        (−7)² = 3 ( mod 23)

        ⇒ ( $-7^{4}$) = 9 (mod 23)

        ⇒ ($-7^{8}$) = 12 (mod 23)

        ⇒ (1$6^{11}$ ) =  1 (mod 23)

        ⇒ 1$6^{99}$ = 1 (mod 23)       ....................(II)

• From (I) and (II)

       1$6^{99}$ = 1 (mod 19x23)  = 1 (mod 437)

       ⇒ 1$6^{99}$ - 1  = 0 (mod 437)

Hence Proved.

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