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I think you are supposed to use Fermat's little theorem but I don't know how to apply it. I tried using exponent laws but I don't know about any such laws that work in this scenario.

zwim
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    Edit: I flattened the exponents in title, we try to keep titles not too vertically demanding. – zwim Mar 22 '20 at 18:17

1 Answers1

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$$F=3^{15^{2n+1}}\equiv3^{15^{2n+1}\pmod{28}}\pmod{29}$$

Now $15^2\equiv1\pmod{28}$

$$\implies F\equiv3^{15(1)^n}\pmod{29}$$

Now $3^3\equiv-2\pmod{29}$

$\implies3^{15}=(3^3)^5\equiv(-2)^5\equiv-3\equiv-3+29$

lab bhattacharjee
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  • i.e. Fermat $\Rightarrow 3^{\large\color{#0a0}{28}}!\equiv1\pmod{!29},$ so $,3^{\large 15^{\Large N}}!!!\equiv 3^{\large 15^{\Large N}!\bmod \color{#0a0}{28}},$ by modular order reduction. By CCRT $,15^{\large\color{#c00} 2}\equiv 1\pmod{!28}^{\phantom{|^{|^|}}}!!$ by $,15^{\large\color{#c00}2}!\equiv (\pm1)^2\equiv1 \pmod{!4\ &\ 7},,$ so $,15^{N}!\equiv 15^{\large N\bmod\color{#c00}2}\pmod{!28}\ \ $ – Bill Dubuque Mar 22 '20 at 19:36