Based on euler theorem, i know that if gcd(a,p)=1 then a^(p-1) congruent to 1 mod p. is (p-1) the smaller value such that this is true? Im wondering if it is possible that a^(q-1) is congruent to 1 mod p where (q-1)<p-1 and gcd(a,q)=1

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- May 12th 2012, 04:13 PMalexandrabel90euler theorem
Based on euler theorem, i know that if gcd(a,p)=1 then a^(p-1) congruent to 1 mod p. is (p-1) the smaller value such that this is true? Im wondering if it is possible that a^(q-1) is congruent to 1 mod p where (q-1)<p-1 and gcd(a,q)=1

- May 13th 2012, 09:22 AMSylvia104Re: Euler theorem
It is certainly possible. Just take $\displaystyle a=1.$