How do I prove that (5k+1)^5^n is congruent to 1 (mod 5^n+1) ? Can I deduce from this result that no number congruent to 1 mod 5 is a primitive root mod 5^m, for any m greater than or equal to 1?
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Originally Posted by Cairo How do I prove that (5k+1)^5^n is congruent to 1 (mod 5^n+1) ? Can I deduce from this result that no number congruent to 1 mod 5 is a primitive root mod 5^m, for any m greater than or equal to 1? and as is divisible by (proof...?) , we are then done. Tonio
isn't divisible by for . But anyway, the congruence is mod .
Originally Posted by HappyJoe isn't divisible by for . But anyway, the congruence is mod . Of course: the sum should have been split into three parts: for Tonio
Sorry, the congruence is modulo 5^(n+1). I've tried using induction to prove the result, but got lost.
Originally Posted by Cairo Sorry, the congruence is modulo 5^(n+1). I've tried using induction to prove the result, but got lost. Tonio's post still applies.
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