Solving $7^x\bmod {29} = 23 $ I have $$7^x\bmod {29} = 23 $$
It is possible to get $x$ by trying out different numbers but that will not be possible if $x$ is actually big.
Are there any other solutions for this equation?
Kind regards
 A: Trial and error is your only option for this type of problem.
The general problem is called the discrete logarithm, and it is hard.
Give prime $p$ and $a,b$ not divisible by $p,$ finding an integer $x$ so that:
$$a^x\equiv b\pmod p$$
might not even have a solution. There is a solution if and only if $o(b)\mid o(a),$ where $o(c)$ is the multiplicative order of $c,$ the smallest $k>0$ such that $c^k\equiv 1\pmod p.$
But even computing $o(a)$ is non-trivial, unless you can quickly list the divisors of $p-1.$
A: I don't think there is anything but trial and error.
But look.  $7^2 \equiv 23 \equiv -6 \pmod {29}$
$7^{2x} \equiv 36 \equiv 7\pmod {29}$.
So $7^{2x-1} \equiv 1 \pmod {29}$.
We know by FLT the $7^{28}\equiv 1$ so for the least power to which $7$ is a multiple of $1$ must be a divisor of $28$.
So $2x-1$ is a multiple of a divisor of $28$.  Immediately we have $2x-1$ is odd so it must be a multiple of an odd divisor of $28$.
The only nontrivial odd divisor of $28$ is $7$.  So test to see if $7^7\equiv 1 \pmod {29}$.  We have $7^7 \equiv 1\pmod {29}$. That was lucky. so we have $2x-1 = 7k$ and  as there are only $7$ values of $7^w\pmod {29}$, we might as well assume $x \le 6$. so $x = 4$ is only option.  Let's try that.
$7^4 \equiv 23\pmod {29}$.
Yep.... that worked.
A: your question $7^x\pmod {29} = 23$ can be written as $7^x$ - $29$ $\lfloor \frac {7^x}{29} \rfloor$ = $23$. Maybe you can proceed with this or maybe the most efficient way to solve will be to use trial and error.
