Question

Let $\mathbf{MODEXP}=\{\mathbf{ha},\mathbf{b},\mathbf{c},\mathbf{pi}|\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{andp}$are positive binary integers such that $\mathbf{ab}\equiv \mathbf{c}\left(\mathbf{modp}\right)\}.$

Show that $\mathbf{MODEXP}\in \mathbf{P}$. (Note that the most obvious algorithm doesn’t run in polynomial time. Hint: Try it first where b is a power of $\mathbf{2}$ .)

Short answer

The solution is,

$MODEXP=\left\{\begin{array}{l}a,b,c,p|a,b,candparebinaryintegers\\ suchthat{a}^b=c(modp)\end{array}\right.$

Step-by-step solution

To Comparable Modulo

Throughout this segment, all of the numbers represent integers. If two integers differ by a multiple of$pforanyplargerthan1$, we say they are comparable modulo$p$.

The $xy$if the integers $xandy$are equal modulo $p\left(modp\right)$We'll call the lowest nonnegative $yxmodp,wherexy\left(modp\right).$Every number modulo $p$is comparable to a member of the set $Zp=0,...,p1$.

To Algorithmmodulo Test

Whenever we state that p satisfies with Fermat test at a, we imply that $a^{P-1}$$\left(modp\right)$

Assuming $p$ satisfies the Fermat test at such $a$ , we can get square roots of $1$since$a^{P-1}$$\mathrm{mod} p=1$, and so$a^{P-1}$$a\left(p_1\right)/2modp$ is a square root of $1$.

They prove first that maybe if $p$his prime, there is no witness, and hence no branch of the algorithm rejects $\left(a{p}_{1}modp\right)6=1$. if a were a stage$4$ witness, and Fermat's little theorem predicts that $p$ is composite