I want to solve this modular equation:
$ 7x + 5 \equiv 2^{11^{2017}} \pmod {31}$
As far as I know, dividing the exponent by 31 and substituting it with the remainder is not allowed.
I've looked here and here but I don't see how to apply that to my equation. I'm sorry for this short question but I don't really know how to approach this equation.
$2$ has order $5\bmod 31$, so $$2^{11^{2017}}\equiv2^{11^{2017}\bmod 5}\pmod{31}.$$ Now $11\equiv 1\mod 5$, so $11^{2017}\equiv 1\mod 5$ and eventually $2^{11^{2017}}\equiv2^1=2\mod{31}$, so the congruence equation becomes $$7x+5\equiv 2\iff7x\equiv -3\iff x\equiv 9(-3)=-27\equiv \color{red}{4}\mod31$$ since $7^{-1}\equiv 9\mod 31$.
$\!\bmod 31\!:\ 2^{\large\color{#c00} 5}\equiv 1\,\Rightarrow\, 2^{\large 11^{\Large n}}\!\! \equiv 2^{\large 11^{\Large n}\bmod\color{#c00}5}\!\equiv 2\ $ by $\ 11^{\large n}\equiv 1^{\large n}\equiv 1\pmod{\!\color{#c00}5},\, $ by Lemma below.
Thus $\,7 x + 5\equiv 2\iff x\equiv \dfrac{-3}{7}\equiv \dfrac{-3\cdot 4}{\,\ \ 7\cdot 4}\equiv \dfrac{-12}{-3}\equiv 4\ $ by Gauss's algorithm.
Lemma $\ $ If $\,\ \color{#c00}{a^{\large n}\equiv 1}\,\ $ then $\ a^{\large K}\equiv a^{\large k}\ $ if $\,K\equiv k\pmod n $
Proof $\ $ Wlog $\,K\ge k\,$ so $\, K = k + jn\ $ for $\ j\ge 0.\ $ Therefore
$$ a^{\large K}\equiv a^{\large k+jn}\equiv a^{\large k} (\color{#c00}{a^{\large n}})^{\large j}\equiv a^{\large k}\, \color{#c00}{1}^{\large j}\equiv a^k $$
using basic congruence rules such as the Congruence Power Rule.
As far as I know, dividing the exponent by 31 and substituting it with the remainder is not allowed.
Correct. $a^k \not \equiv a^{(k\mod n)} \mod n$ is certainly not allowed.
(Although modulo distributes over addition and multiplication; $(a + \text{multiple of }n)\circ(b + \text{multiple of }n)= a\circ b + \text{multiple of }n$ when $\circ$ is either $+$ or $\times$--- it clearly does not distribute over exponents; $a^{k+ \text{multiple of }n} = a^k\times a^{\text{multiple of }n} \ne a^k + \text{multiple of }n$-- it simply does not.)
However you are allowed the two following.
if $a \equiv b \pmod n$ then $a^k \equiv b^k \pmod n$--
and if $a^m\equiv 1 \pmod n$ then $a^{(k\pmod m)} \equiv a^k \mod n$.
This is because $a^{k + m*h} = a^k*(a^m)^h \equiv a^k*(1)^h \equiv a^k \mod n$.
.....
Thus if we know $2^m \equiv 1 \pmod {31}$
Then $2^{11^{2017} } \equiv 2^{(11 \pmod m)^{2017}} \pmod {31}$.
So if we know $2^m \equiv 1 \pmod {31}$. And we know $11 \equiv a \mod m$. And we know $a^k \equiv 1 \pmod m$. And we know $2017 \equiv b \pmod k$ the we know:
$2^{11^{2017}}\equiv 2^{(11^{2017}\pmod m)} \equiv$
$2^{(11\pmod m)^{2017}}\equiv 2^{a^{2017}} \equiv 2^{a^{2017 \mod k}}$
$2^{a^b} \pmod {31}$.
.....
Which is not as abstract as it looks.
Why know by Fermat's Little Theorem that $2^{30} \equiv 1 \pmod {31}$ and by experimenting with smaller factors of $30$ we find that $2^5 = 32 \equiv 1 \pmod {31}$.
So $2^{11^{2017}}\equiv 2^{(11^{2017} \mod 5)}\equiv 1 \pmod {31}$.
Now clearly $11^{2017} \equiv 1^{2017} \equiv 1 \pmod 5$ so
$2^{11^{2017}}\equiv 2^1 \equiv 2 \pmod {31}$.
....
To put it another way.
$2^{11^{2017}} = 2^{(1 + 10)^{2017}} = 2^{1 + \text{some multiple of } 10}=$
$2*2^{\text{some multiple of } 10} = 2*((2^5)^2)^{\text{some integer}}=$
$2*(32)^{2*\text{some integer}} \equiv 2*1^{2*\text{some integer}}\pmod {32} \equiv 2\pmod {31}$.
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Oh, to actually solve the original question:
$7x + 5 \equiv 2^{11^{2017}} \pmod {31}$
$7x + 5 \equiv 2 \pmod {31}$
$7x \equiv -3 \pmod {31}$
Well, we could get lucky and notice $7x \equiv -3 \equiv 28\pmod {31}$ so $x \equiv 4 \pmod {31}$ will be one set of solutions. But is it the only one? And could we have solved it without getting lucky?
Well, we know if $\gcd (n,k) =1$ then we know that there exist $x, m$ so that $kx + mn = 1$.... or in other words thatn $kx \equiv 1 \mod n$ has a solution.
And if $n$ is prime we know that the solutions to $kx\equiv 1\mod n$ is unique up to modulo $31$ equivalence.
So $7m\equiv 1 \pmod {31}$ has a unique solution. By brute force/luck we can find that $7*9 = 63 \equiv 1 \pmod {31}$.
So $9$ is the multiplicative inverse of $7$ in tems of modulo $31$ equivalence.
So $7x + 5 \equiv 2 \pmod (31)$
$7x \equiv -3\pmod (31)$
$9*7x \equiv -3*9 \pmod(31)$
$63x \equiv -27 \pmod (31)$
$x \equiv 4 \pmod(31)$.
To solve such question, I would simplify the right hand side first. Notice that ${(2^{11})}^{2017} \equiv 2^{2017} \pmod{31}$. Further more, $2017 = 67 \times 30 + 7$. Therefore, By Fermat's theorem, $2^{2017} \equiv {(2^{30})}^{67} \times 2^7 \equiv 1 \times 2^7 \equiv 128 \equiv 4 \pmod{31}$.
The equation finally becomes
$$7x+5 \equiv 4 \pmod{31} \to 7x \equiv -1 \pmod{31}$$
Notice that I divided by $30$ instead of $31$ when using Fermat little theorem.
$$2^5\equiv1(\mod31).$$ Thus, $$2^{11^{2017}}=2^{11\left(121^{1008}-1\right)+11}\equiv2^{11}\equiv2.$$ Id est we need to solve $$7x\equiv-3(\mod31)$$ or $$63x\equiv-27(\mod31)$$ or $$x\equiv4(\mod31).$$
