Applying Fermat's Little Theorem to show $7^{8n+3}+2 = 5p$. I just read about Fermats little theorem and was wondering if the following relationship is an extension of this:
$7^{8n+3}+2 = 5p$ where $p$ is an real integer.
If so can you show me how/why this is?
 A: It is not an extension but a consequence of it. 
You assert that 
$7^{8n+3}+2 \equiv 0 \mod{5}$.
So $2^{8n+3} \equiv -2 \mod{5}$. 
Which multiplying by $2$ is equivalent to 
$2^{8n+4} \equiv -4 = 1 \mod{5}$ 
By Fermat's Little Theorem 
$2^4 \equiv 1 \mod{5}$ and so $2^{8n+4}= (2^4)^{2n+1}$ is indeed $1$ modulo $5$.
A: Hint $\ {\rm mod}\ 5\!:\,\ \overbrace{\color{#c00}7^{\large 3+8n}\equiv\, \color{#c00}2^{\large 3+8n}}^{\color{#c00}{\large 7\,\equiv\,2}}\equiv\, \overbrace{2^{\large 3} (\color{#b0f}{2^{\large 4}})^{\large 2n}\equiv\, 8\color{#b0f}{(1)}^{\large 2n}}^{\color{#b0f}{\large 2^4\,\equiv\, 1}}\equiv\, \color{#0a0}3\,$
So adding $\,2\,$ to it yields $\,2+\color{#0a0}3\equiv 0.\,$ Here we used standard Congruence Rules.
Key Idea $ $ Generally, if $\,a^e\equiv 1\pmod m\,$ then exponents on $\,a\,$ can be taken mod $\,e,\,$ i.e. $\ a^{\large j}\equiv a^{\large k}\pmod m\,\ $ if $\,\ j\equiv k\pmod e.\ $ This may be proved exactly as above, i.e.
$$ \begin{array}{}\color{#b0f}{a^{\large e}\equiv 1}\\ j = k\! +\! en\end{array}\Rightarrow\,\ a^{\large j}\equiv a^{\large  k+en}\equiv a^{\large  k}\color{#b0f}{(a^{\large  e})}^{\large  n}\equiv a^{\large  k}\color{#b0f}{(1)}^{\large  n}\equiv a^k\!\!\pmod m\qquad $$
So in your problem, because $\, 2^{\large \color{#b0f}4}\equiv 1\pmod{5},\,$ exponents on $\,2\,$ can be taken mod $\,\color{#b04}4.\,$ Thus  $\,3\!+\!8n\equiv 3\pmod{\color{#b0f}4},\,$ so $\,2^{\large 3+8n}\equiv 2^{\large 3}\pmod 5$
So it is not an extension of Fermat's theorem but, rather, a consequence of periodicity of powers i.e. $\,a^e\equiv 1\,\Rightarrow\, a^{en}\equiv 1,\,$ as in the above proof.
