What is the sum of all $\alpha$ such that $5\mid(2^\alpha+\alpha)(3^\alpha+\alpha)$ and $\alpha$ is a positive whole number less than $120$? What is the sum of all $\alpha$ such that $5\mid(2^\alpha+\alpha)(3^\alpha+\alpha)$ and $\alpha$ is a positive whole number less than $120$?
For context, this is a problem from a friend that we have both not been able to figure out.
Here's what I've tried: 
$(2^\alpha+\alpha)(3^\alpha+\alpha)\equiv0\pmod{5}$,
$6^\alpha+\alpha(2^\alpha)+\alpha(3^\alpha)+\alpha^2\equiv0\pmod{5}$,
$1^\alpha+\alpha(2^\alpha)+\alpha(3^\alpha)+\alpha^2\equiv0\pmod{5}$, 
$\alpha(2^\alpha+3^\alpha+\alpha)\equiv4\pmod{5}.$
I got stuck here and I'm not sure what to do now. Does anyone have any suggestions or a new method I could try?
*also I would prefer a hint rather than a solution. Thanks!
 A: Below we explain the key idea of the general method  (stop at that paragraph for your hint). .
Notice $\,\ 5\mid (2^n+n)(3^n+n)\iff 5\mid 2^n+n\,$ or $\,5\mid 3^n+n,\,$ by $5$ prime, hence we reduce to solving $\,5\mid a^n+n =: f(n),\,$ for $\,a\in\{2,3\}$.
Key Idea $ $ We eliminate nasty exponential dependence on $\,n\,$ in $\,a^{\large n}\,$ by noting - by Fermat - that it is periodic $\!\bmod 5\!:\ a^{\large\color{darkorange} 4}\equiv 1\,\Rightarrow\, a^{\large n}\! =a^{\large i+\color{darkorange}4k}\equiv \color{90f}{a^{\large i}}\,$ by mod $\small\rm\color{darkorange}{exp}$onent reduction. So we get a simpler linear $ $ congruence by replacing $ $ function $\, g(n) = \color{#90f}{a^{\large n}}\,$ by constants $\,\color{90f}{a^{\Large i}},\ i = 0,1,2,3$.
Doing so $\ n = \color{}i\!+\!4\:\!\color{#c00}k\,\Rightarrow$ $\! \bmod\color{#c00} 5\!:\ \overbrace{0\equiv f(i\!+\!4k)\equiv \color{90f}{a^{\Large\color{}i}}\!+\! i\!+\!4k}^{\Large{\ \  5\ \, \mid\, \  f(n)\ \ =\ \  \color{#90f}{a^{\LARGE n}}\ + \  n}} \equiv f(i)\!-\!k\iff \color{#c00}{k\equiv f(i)}$
therefore $\ n= i\!+\!4\color{#c00}k \,=\, i+ 4(\color{#c00}{f(i)\!+\!5}j)\,\equiv\ \bbox[5px,border:1px solid #0a0]{i\!+\!4f(i)}^{\phantom |}\pmod{\!20}$
so $\,\ \ i\!=\!0\Rightarrow n\equiv 0\!+\!4f(0)\equiv 0+4\{\ 1,\,\ \ 1\}\equiv\bbox[5px,border:1px solid #0a0]{\ \ 4,\,\ 4}\ $ for $\,a\in\{2,3\}$
$\quad\ \ \ i\!=\!1\:\!\Rightarrow n^{\phantom{|^I}}\!\!\!\!\equiv 1\!+\!4f(1)\equiv1+4\{\ 3,\,\ \ 4\}\equiv \bbox[5px,border:1px solid #0a0]{13,17}\,$
$\quad\ \ \ i\!=\!2\:\!\Rightarrow n^{\phantom{|^I}}\!\!\!\!\equiv 2\!+\!4f(2)\equiv2+4\{\ 6,\ 11\}\equiv \bbox[5px,border:1px solid #0a0]{\:\! \ 6, \ 6}\,$
$\quad\ \ \ i\!=\!3\:\!\Rightarrow n^{\phantom{|^I}}\!\!\!\!\equiv 3\!+\!4f(3)\equiv3+4\{11,10\}\equiv \bbox[5px,border:1px solid #0a0]{\ \ \:\! 7,\ 3}\,$
So we conclude the solutions are $\ n\equiv \bbox[5px,border:1px solid #0a0]{3,4,6,7,13,17}\,\pmod{\!20}.\,$ Note each solution extends to $\,6\,$ solutions below $120\!:\ n+20k,\, k = 0,1,2\ldots,5,\,$ so now it is easy to compute the sought sum.
Remark $ $ The same idea works generally. If $\,g(n)\,$ has period $\,\ell,\,$  so $\ g(i+ \ell k) = g(i),\,$ then we can solve $\,f(n,g(n)) \equiv a\pmod{\!m}\,$ by using this periodicity to eliminate the function $\,g(n)\,$ as above -see here for further detail, which includes another worked example.
