SOLUTION: Show that if am b, k and m are integers such that k≥1, m≥2, and a≡b( mod m), then a^k≡b^k( mod m)

Question 1074892: Show that if am b, k and m are integers such that k≥1, m≥2, and a≡b( mod m), then a^k≡b^k( mod m)
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
a ≡ b( mod m ) means that a - b = cm for some integer c
:
a = b + cm
:
We use the binomial theorem
:
a^k = (b + cm)^k = b^k + k*b^(k-1)*cm + ......
:
Note that every term on the right except b^k is a multiple of m because it contains a power of cm
:
therefore
:
*************************************************
a^k - b^k = multiple of m so a^k congruent to b^k
*************************************************
:
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SOLUTION: Show that if am b, k and m are integers such that k&#8805;1, m&#8805;2, and a&#8801;b( mod m), then a^k&#8801;b^k( mod m)

SOLUTION: Show that if am b, k and m are integers such that k≥1, m≥2, and a≡b( mod m), then a^k≡b^k( mod m)