Full-period criterion for a mixed congruential generator ($\mathrm{period}=M$)

A linear congruential generator has period $M$ for every seed if and only if $\gcd (c,M)=1$, every prime dividing $M$ divides $a-1$, and $4\mid M$ implies $4\mid a-1$. The coprimality of $c$ prevents the orbit from being trapped in a residue class; the multiplier conditions permit full-period lifting through each prime power. Combining the prime-power periods by the Chinese remainder theorem gives period $M$. For $M=12^k=2^{2k}{3}^k$, the multiplier criterion reduces to $a\equiv 1(\mathrm{mod}12)$, together with $\gcd (c,6)=1$. A positive increment alone is insufficient.