Decrypt 01 09 00 12 12 09 24 10 knowing encryption was doing with c=character $c^5\pmod{29}$

Question

So the solution said in order to find out the original character number($A=0, B=1, C=2$, etc). That you have to find $5x\equiv1\pmod{28}$ so $x\equiv17\pmod{29}$, because $\phi(29)=28$. So the solution becomes $d$=encrypted word then c=$d^{17}\pmod{29}$ How does getting the $\phi$ give the equation for decryption when $e=5$ and $p=29$ in the equation $c^5\pmod{29}$?

Answer 1

I agree that $\phi(29)=28$, as $29$ is prime. The encryption exponent is $5$ and so the decryption exponent is its inverse modulo $28$, which indeed is $17$.

So the decryption function is $x \to x^{17} \pmod{29}$. The inverse of an exponential function is another exponential function, over such a finite ring. The inverse of $e$ modulo $\phi(n)$ is the required exponent $d$.

So $1,9,0,12,12,9,24,10$ decrypted becomes

$1,4,0,12,12,4,20,15$ which is "beammeup" so "beam me up" (with 29 characters a space would have been nice...)