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Old 2020-08-02, 17:20   #5
amenezes
 
Aug 2020

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In the Theorem proof the error is here:
2^(n-1)==1 ( mod n) = 1 (mod p) as p | n
b^(n-1)==1( mod n) == 1 (mod p) as p | n
Subtracting the two equivalences gives:
2^(n-1)==b^(n-1) ( mod p)
Taking the square roots of both sides gives:
2((n-1)/2)==+/- b^((n-1)/2)( mod p)
and not
2^((n-1)/2) == b^((n-1)/2) (mod p)
as claimed in the proof.
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