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disclaimer : I am new to Congruences

I have the Following Congruence, and I am a bit stuck when trying to solve it

$2^{2^{(6L+2)}} + 11 \equiv 8 (mod19)$

what i have done is the following:

$16^{2^{(6L)}} \equiv 16 (mod19)$

$16^{64^{(L)}} \equiv 16 (mod19)$

at this stage I am stuck, what can I do now ?

Eitank
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    One needs to know the concept of the (multiplicative) order of an element in modular arithmetic: $16^k \equiv 1\pmod{19}$ if and only if $k$ is a multiple of the order of $16$ modulo $19$. – Greg Martin Dec 28 '20 at 07:17
  • @GregMartin, can you please elaborate more. i really want to understand this – Eitank Dec 28 '20 at 10:48
  • @GregMartin, i have read about multiplicative order and i saw that for 16 it is 9. because $16^9 \equiv 1 (mod19)$ – Eitank Dec 28 '20 at 10:50
  • $\bmod 18!:\ 64\equiv 10,,$ and $10^{2}\equiv 10\Rightarrow \color{#c00}{10^{L}\equiv 10},$ for all $,L>0,$ (cf. idempotents here or here), so $,2^{2+6L}! = 4\cdot 64^L\equiv 4(\color{#c00}{10^L})\equiv 4(\color{#c00}{10})\equiv \color{#0a0}4,,$ both for $,L>0,,$ and $,L= 0,$ (by $2^{2+6\cdot 0} = 4)\ \ $ – Bill Dubuque Dec 28 '20 at 15:00
  • Thus for all $,L\ge 0,$ we have $\bmod 19!:\ 2^{\large 2^{2+6L}}!!\equiv 2^{\large\color{#0a0}4}\equiv 16,$ by little Fermat & modular order reduction $ \ \ $ – Bill Dubuque Dec 28 '20 at 15:00

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