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)
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-> 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)
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You can put this solution on YOUR website! a ≡ b( mod m ) means that a - b = cm for some integer c
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a = b + cm
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We use the binomial theorem
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a^k = (b + cm)^k = b^k + k*b^(k-1)*cm + ......
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Note that every term on the right except b^k is a multiple of m because it contains a power of cm
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therefore
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a^k - b^k = multiple of m so a^k congruent to b^k
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