Proving $x\equiv 2^{11} \mod{97}$

Question

In $\mathbb{N}$ we suppose the equation $(F): x^{35} \equiv 2 \mod{97}$, and $(97,x)=1$.

How could I prove that $x\equiv 2^{11} \mod{97}$ ?

Answer 1

From little Fermat, since $(x,97) = 1$, we have that $$x^{96} \equiv 1 \pmod{97}$$ We are given that $x^{35} \equiv 2 \pmod{97}$. Now $$(x^{35})^{11} \pmod{97} = x^{385} \pmod{97} = x^{4 \times 96} \cdot x \pmod{97} \equiv x \pmod{97}$$ Hence, $$x \equiv (x^{35})^{11} \pmod{97} \equiv 2^{11} \pmod{97}$$

Answer 2

Raise the $\rm\color{#0A0}{equation}$ to the $\rm\!\color{#C00}{{\bf\,11}'th\ power},\ $ use $\rm\ mod\,\ 97\!:\ \color{#c1f}{x^{\large96}\!\equiv 1}\,\ if\,\ x\not\equiv 0\,$ by Fermat, $\,97\,$ prime.

$\rm\ \ mod\ \ 97\!: \ \color{#0A0}{2\equiv x^{\large 35}}\,\Rightarrow\, \color{#0A0}2^{\color{#C00}{\large \bf 11}}\!\equiv (\color{#0A0}{x^{\large 35}})^{\color{#C00}{\large \bf 11}}\!\equiv x^{\large\color{blue}{1+96\,N}}\!\equiv x\,\color{#C1f}{(x^{\large 96})^{\large N}}\! \equiv\, x\color{#C1f}{(1)^{\large N}}\! \equiv\, x,\ $ because

$\rm\ \ \color{#0A0}{35}\cdot\color{#C00}{11} =\color{blue}{1\!+\!96N}\ $ by $\rm\:mod\ 96\!:\ \color{#0A0}{35}\cdot\color{#C00}{11} \equiv 35(3\cdot 3+2)\,\equiv\, 105\cdot3+70\,\equiv\, 9\cdot 3+70\,\equiv\, 1$

Answer 3

I would like to show how $11$ is identified.

Let $x\equiv2^y\pmod{97}\implies (2^y)^{35}\equiv2 \pmod{97}$

$\implies 2^{35y-1}\equiv1\pmod {97}$ (dividing either sides by $2$ as $(2,97)=1$ )

Using Fermat's Little Theorem, $2^{97-1}\equiv1\pmod {97}$

So, it is sufficient to find $y$ such that $96$ divides $(35y-1)$ i.e., there will be some integer $z$ such that $35y-1=96z\iff 35y-96z=1$

Expressing $\frac{96}{35}$ as continued fraction,

$$\frac{96}{35}=2+\frac{26}{35}=2+\frac1{\frac{35}{26}}=2+\frac1{1+\frac9{26}}$$ $$=2+\frac1{1+\frac1{\frac{26}9}}=2+\frac1{1+\frac1{2+\frac89}}=2+\frac1{1+\frac1{2+\frac1{\frac98}}}$$ $$=2+\frac1{1+\frac1{2+\frac1{1+\frac18}}}$$

So, the last convergent of $\frac{96}{35}$ is $$2+\frac1{1+\frac1{2+\frac11}}=\frac{11}4$$

Using convergent property of continued fraction, $$96\cdot4-35\cdot11=-1$$

So, $35y-96z=-(96\cdot4-35\cdot11)\implies \frac{35(y-11)}{96}=z-4$ which is an integer

$\implies 96$ divides $35(y-11)$

$\implies 96$ divides $(y-11)$ as $(35,96)=1$

So, $y\equiv11\pmod{96}$

Answer 4

Here we look for relations ('tricks') that will solve the problem without using Fermat's Little Theorem.

We interpret the OP's question as a 'how to calculate' ${2^{11}}^{35} \pmod{97}$.

Well $2^{11} \equiv 2^8 \times 2^3 \equiv 62 \times 8 \equiv 11 \pmod{97}$ so we've found our first relation,

$\tag 1 {2}^{11} {11}^{-1} \equiv 1 \pmod{97}$

We suspect that progress can be made by calculating ${11}^{-1} \pmod{97}$, i.e. solving the equation

$\quad 11x \equiv 1 \pmod{97}$

This can be done by hand using an algorithm discussed here. Proceeding,

Multiply by $ 9 $:

$\quad 99 x \equiv 9 \pmod{97}$
$\quad 2 x \equiv 9 \pmod{97}$

Multiply by $ 48 $:

$\quad 96 x \equiv 432 \pmod{97}$
$\quad -1 x \equiv 44 \pmod{97}$
$\quad 1 x \equiv 53 \pmod{97}$

So

$\quad 11^{-1} \equiv 53 \pmod{97}$

But then

$\quad 2^{1}11^{-1} \equiv 9 \pmod{97}$
$\quad 2^{2}11^{-2} \equiv 81 \equiv -16 \equiv - 2^4 \pmod{97}$

and we can write down another relation,

$\tag 2 2^{-2}11^{-2} \equiv -1 \pmod{97}$

The two (primitive) relations are all we need.

$\quad \displaystyle {2^{11}}^{35} \equiv 11^{35} \equiv 11^{35} (-1) (11^{-34}) (2^{-34}) \equiv (-1) 11 \,{(2^{11})}^{-3}\, 2^{-1} \equiv (-1) \,({11}^{-2}) \, (2^{-1}) \equiv$
$\quad \quad (-1) \,\big [ ({11}^{-2}) \, (2^{-2}) \big ] \,2 \equiv (-1) \,(-1) \,2 \equiv 2 \pmod{97}$