<s>[INST] How does one get from $p+3=3k+2$ then $2^{p+3} \equiv 4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$ How does one get from $p+3=3k+2$ then $2^{p+3}\equiv4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$.
I am just starting with modular arithmetic so any help would be greatly appreaciated.
Credit to user236182 for original answer, just wanted to know how he got to it
 [/INST] First look at the powers of $2$ modulo $7$. In particular note that $2^3\equiv1\pmod7$. Hence $2^{3k+a}\equiv 2^{3k}2^a\equiv2^a\pmod7$.
This gives us the first part of the question:
$$p+3=3k+2\hspace{5mm}\implies\hspace{5mm} 2^{p+3}\equiv2^{3k+2}\equiv2^2\equiv4\pmod7$$
The second part is pretty straightforward too:
$$5\cdot4-31\equiv-11\equiv3\pmod 7$$
hence
$$5\cdot2^{p+3}-31\equiv3\pmod7$$
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