I simplified the problem as writing it first as $2^{2^{6n+2}}+3\equiv 0(mod 19)$ finally ending till $2^{2^{6n+2}}\equiv 2^4(mod 19)$.I cannot figure out how to deal with the exponents with the same base on both sides of the congruence.
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4So, what have you tried so far? – Paresseux Nguyen Jun 03 '21 at 05:02
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1Have you tried to find the multiplicative order of $2$ in $pmod{19}$? – daruma Jun 03 '21 at 05:24
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what is $x?$ And add your work to the question. not just in comments. – Thomas Andrews Jun 03 '21 at 05:24
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Hint: $,2^{\large N}!\equiv 2^{\large N\bmod\color{#c00}{18}}!\pmod{!19},,$ and $,2^{\large 6n}!\equiv 1\pmod{!9}\overset{\times\ 4}\Longrightarrow 4\cdot 2^{\large 6n}\equiv 4\pmod{\color{#c00}{!!18}}\ \ $ – Bill Dubuque Jun 03 '21 at 05:42
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Or, operationally $\ 4\cdot 2^{6n}\bmod 18 = 2(2\cdot 2^{6n}\bmod 9) = 2(2\cdot 1^n) = 4\ $ via the mod Distributive Law – Bill Dubuque Jun 03 '21 at 05:52
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You just need to show that $ 16 \mid 2^{2^{6n+2} }$. To that end, write it as $ (2^{2^{6n}})^4 $, and simply observe $ 2 \mid 2^{2^{6n}} $. – Hello Jun 03 '21 at 06:00
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By Fermat's little theorem,
$2^{18}\equiv 1\pmod{19}$.
So $2^{18k+4}\equiv 2^4\equiv -3\pmod{19}$ for any integer $k$. You can check that $18$ is the multiplicative order of $2$ although actually we don't need it.
Now, we want to find the residue of $4^{3n+1}$ modulo $18$.
$4^{3n+1}\equiv 64^n\cdot 4\equiv 10^n\cdot 4\pmod{18}$
You can check regardless of whether $n$ is even or odd, $4^{3n+1}\equiv 4\pmod{18}$.
So $2^{4^{3n+1}}+3\equiv 2^{18k+4}+3\equiv 2^4+3\equiv 0\pmod{19}$.
daruma
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2Please strive not to add more dupe answers to dupes of FAQs, cf. site policy here. – Bill Dubuque Jun 03 '21 at 05:54