Exponent reduction for units ($a^k\equiv a^{k\mathrm{mod}\phi (n)}(\mathrm{mod}n)$)

For a unit modulo n $a$ and a nonnegative integer $k$, division with remainder writes $k=q\phi (n)+r$ with $0\le r<\phi (n)$. Euler's theorem then givesThis replaces a potentially enormous exponent by a small remainder. The exponent itself is reduced modulo Euler's totient function, while the powers are reduced modulo $n$: these are different integer moduli.

For a worked modulo-fifteen calculation, seven is coprime to fifteen and $\phi (15)=8$. ThereforeUse exponentiation by squaring on the remaining exponent $5=(101{)}_2$, reducing after each multiplication:Thus $7^{2029}\mathrm{mod}15=7$. Modular exponentiation supplies the computation; Euler's theorem supplies the optional shortening of its exponent. Exponentiation by squaring also works for nonunit bases, without using this shortening.

The guaranteed exponent $\phi (n)$ is a common return exponent and an upper bound on the shortest period; it need not be that period. The multiplicative order $d={\mathrm{ord}}_n(a)$ is the least positive return exponent for the chosen unit. The Carmichael function $\lambda (n)$ is the least positive return exponent that works for every unit. Their precise relation isEither smaller return exponent can replace $\phi (n)$ in exponent reduction for units. Modulo fifteen, ${\mathrm{ord}}_{15}(7)=\lambda (15)=4<8=\phi (15)$, so $2029=507\cdot 4+1$ gives $7^{2029}\equiv 7$ immediately. For base four the multiplicative order is only two, smaller even than $\lambda (15)$. A return at exponent eight alone does not prove that the multiplicative order is eight.^[ref]

If the reduced exponent is zero, the rule gives $a^k\equiv a^0=1$, and still requires a unit. Applying it to the nonunit three would falsely predict $3^8\equiv 3^0=1(\mathrm{mod}15)$; the actual residue is six. For negative exponents, first interpret the power using a modular multiplicative inverse, as in modular exponentiation.