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3 added 49 characters in body

By Fermat's little theorem we know that

$$b^{p-1}=1 \mod p$$

if p is prime and $\gcd(b,p)=1$. On the other hand, I was wondering whether

$$b^{n-1}=-1 \mod n$$

can occur at all?

Update: sorry, I meant n odd. Please excuse.

2 nicer mods

By Fermat's little theorem we know that

$$b^{p-1}=1 \mod p$$

if p is prime and $\gcd(b,p)=1$. On the other hand, I was wondering whether

$$b^{n-1}=-1 \mod n$$

can occur at all?

1

# b^(n-1)=-1 mod n

By Fermat's little theorem we know that

$$b^{p-1}=1 mod p$$

if p is prime and $\gcd(b,p)=1$. On the other hand, I was wondering whether

$$b^{n-1}=-1 mod n$$

can occur at all?