Apparently the general solution of this:

$3=(3388997632\cdot x^{23}) \text{ mod }25$

is $x = 25n + 9$, where $n$ is any natural number, it seems?

I get how there is connection with $25$ as modulo, that is: $x =\text{modulo }n + 9$ but can't see where 9 comes from despite being multiple of 3. I am also seeking general solution to these type of equations, as above...or with variables in letters:

$y = (ax^n)\text{ mod }n+2$

Apparently, sometimes but not always the solution for $x$ is just as above: modulo times any number plus some other number? Also it seems that if $ax^n$ is negative there is no easy "general solution" in other words the general solution doesn't apply here as outlined above. thanks in advance.

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    $\begingroup$ That big number is $=7\pmod{25}$. By Euler's theorem $x^{20}=x^{\phi(25)}=1\pmod{25}$. That means that your equation is the same as $7x^{3}=3\pmod{25}$. Now, $7*18=1\pmod{25}$, and $3*18=4\pmod{25}$. Therefore, the equation is equivalent to $x^3=4\pmod{25}$. You can easily check that $9^3=4\pmod{25}$. $\endgroup$ – user647486 Apr 19 at 14:44
  • $\begingroup$ @thanks for the edit! $\endgroup$ – user3918597 Apr 19 at 14:45
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    $\begingroup$ To deduce that $9$ is the only solution, you can solve first $x^3=4=-1\pmod{5}$, which only requires checking $5$ possible remainders. You get that $x=4\pmod{5}$ is the only solution. Then you search for solutions of this form $5k+4$ of the original equation: $(5k+4)^3=4\pmod{25}$. This reduces to $3\cdot5\cdot4^2k+4^3=4\pmod{25}$, which gives $3\cdot 5\cdot 8\cdot k+5=0\pmod{25}$. This has solution $k=1\pmod{5}$. Therefore, the solutions to the original equation are of the form $5k+4$, with $k=5n+1$. This is, the solutions are of the form $5n+9$. $\endgroup$ – user647486 Apr 19 at 14:56
  • $\begingroup$ ok most is clear but just to clarify: you get 7 from Euler's totient theorem? $\endgroup$ – user3918597 Apr 19 at 14:56
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    $\begingroup$ If your slave isn't handy, note that $100 \equiv 0 \bmod 25$, so you just have to look at the last two digits of $3388997632$. $\endgroup$ – Robert Israel Apr 19 at 15:59

Here's how it connects: $$3\equiv(((33889976\cdot4+1)\cdot25+7)\cdot x^{23}) \pmod{ 25}\implies 3\equiv 2\cdot x^3\pmod 5\\\implies-1\equiv x^3\pmod 5\implies x=5y+4$$ Then applying polynomial remainder theorem, Euler and negative versions to simplify a modular arithmetic fraction, to each case ( we know the base is coprime) we'll see: $$4\equiv 9^{3}\pmod{25}$$ which shows the base to be 9 mod 25.


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