Let $a = 3 + 2\sqrt{2}$ and $b = 3 - 2\sqrt{2}$. Prove that for every natural number $n$, ${a^n}$ + ${b^n}$ is a natural number, not divisible by $5$.
It's trivial that this number is natural by using binomial formula, but I can't figure out how to prove that it's not divisible by $5$.
On
Starting from the hint in the great @lulu comment we have that $$\left(3-2 \sqrt{2}\right)^n+ \left(2 \sqrt{2}+3\right)^n$$ is the solution of the recurrence having characteristic polynomial $$\lambda^2-6\lambda+1$$ generated by the recurrence $$s_n=6s_{n-1}-s_{n-2};\;s_0=2;\;s_1=6$$ If we rewrite the recurrence $\text{mod }5$ we get $$s'_n=s'_{n-1}-s'_{n-2};\;s'_0=2;\;s'_1=1$$ which gives the values $\text{mod }5$ of the recurrence above.
The solution can be easily found by the characteristic equation $$x^2-x+1=0\to x_1=\frac{1-i \sqrt{3}}{2};\;x_2=\frac{1+i \sqrt{3}}{2}$$ so we have $$s'_n=a \left(\frac{1-i \sqrt{3}}{2}\right)^n + b\left(\frac{1+i \sqrt{3}}{2}\right)^n$$ Plugging the initial values $s'_0=2;\;s'_1=1$ we get $$\begin{cases} a+b=2\\ \left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right) a+\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) b=1\\ \end{cases} $$ Solution is $a=1;\;b=1$ therefore the solution of the second recurrence is $$s'_n= \left(\frac{1-i \sqrt{3}}{2}\right)^n + \left(\frac{1+i \sqrt{3}}{2}\right)^n$$ These are the numbers of the set $a^n+b^n$ modulus $5$ and can be easily seen that they are never $0$.
Finally we can say that $a^n+b^n$ is never divisible by $5$.
I am well aware that what follows is not the simplest way to answer to this question, but I'd like to put the problem in a more structured context.
Consider the field $$ \Bbb Q(\sqrt 2)=\{a+b\sqrt2\,\mid\,a, b\in\Bbb Q\}. $$ The field is endowed with an automorphism $\sigma$ defined as $\sigma(a+b\sqrt2)=a-b\sqrt2$ which has the property that $\Bbb Q$ is (the subfield) fixed.
Note that the two elements given are $$ x=3+2\sqrt2\qquad\text{and}\qquad x_\sigma=\sigma(x)=3-2\sqrt2 $$ This tells immediately that $x^n+x_\sigma^n\in\Bbb Q$ since $\sigma(x^n+x_\sigma^n)=x^n+x_\sigma^n$. Then it is pretty clear that it is actually in $\Bbb Z$ since taking positive powers never introduces denominators.
Next observe that $xx^\sigma=1$ so that in fact $$ x^n+x^n_\sigma=\frac{1+x^{2n}}{x^n}\qquad\qquad(\ast) $$ We now want to study divisibility by $5$. Thus we shall work modulo $5$ in $\Bbb Z[\sqrt 2]$. One has $$ \frac{\Bbb Z[\sqrt 2]}{(5)}\simeq\frac{\Bbb Z[T]}{(5,T^2-2)}\simeq \frac{\Bbb F_5[T]}{(T^2-2)}\simeq\Bbb F_{25},\qquad(\ast\ast) $$ the field with $5^2$ elements, since $2$ is not a square modulo $5$. Denote $\bar z$ the class of $z\in\Bbb Z[\sqrt2]$ in this quotient.
Note that $\bar x\neq0$ so that $\bar x^n\neq 0$ for all $n\in\Bbb Z$ so that after $(\ast)$ we need to prove that $\bar x^{2n}\neq-1$.
Now if $u=\bar T$ in $(\ast\ast)$ we can write $\bar x=3+2u$ and a simple modulo $5$ computation shows that $\bar x^2=-\bar x_\sigma$, i.e. $$ \bar x^3=-1. $$ Thus $\bar x^6=1$ and therefore $$ \bar x^k=-1\qquad\Longleftrightarrow\qquad k\equiv3\bmod 6 $$ showing that the numerator of $(\ast)$ is never $0$ modulo $5$.