<s>[INST] Proving $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$ Theorem. $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$.

I'm tasked with proving the given theorem by induction. Here's where I've gotten so far...
Proof. Clearly, the theorem is true for $n=1$, establishing the base case. Moreover, it can easily be shown that the theorem works for $n=2,3,4,5,\ldots, 18$.
Suppose, now, that $k$ is an integer for which the given theorem holds. Consider, then, $k+18$.
$$2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}}$$
By Fermat's Little Theorem, if $n \equiv m \pmod{p-1}$, then $a^m \equiv a^n \pmod{p}$. Clearly, $k+18 \equiv k \pmod{18}$. Then,
$$2^{2^{k+18}} \equiv 2^{2^k} \pmod{19}$$
$$3^{2^{k+18}} \equiv 3^{2^k} \pmod{19}$$
$$5^{2^{k+18}} \equiv 5^{2^k} \pmod{19}$$
Hence, $2^{2^{k+18}}$ may be expressed as $\;2^{2^k} + 19n$ for some integer $n$, $\;3^{2^{k+18}}$ as $\;3^{2^k}+ 19m$ for some integer $m$, and $\;5^{2^{k+18}}$ as $\;5^{2^k} + 19q$ for some integer $q$.
Thus, $2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}} = (2^{2^{k}} + 3^{2^{k}} + 5^{2^{k}}) + 19(m+n+q)$, a sum of integers divisible by $19$, and thus clearly divisible by $19$.
By the principle of induction, we've thus shown that $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$, concluding the proof.
$\blacksquare$

Is this approach valid?
 [/INST] The direct approach uses $-3\equiv 2^4\pmod{19}$ and $-5\equiv 2^{7}\pmod{19}.$
So when $n>0,$ $(-3)^{2^n}=3^{2^n}$ and $(-5)^{2^n}=5^{2^n},$ and we get:
$$2^{2^n}+3^{2^n}+5^{2^n}\equiv 2^{2^n}+\left(2^{2^{n}}\right)^4+\left(2^{2^{n}}\right)^{7} \pmod{19}$$
So the right hand side is $p(2^{2^n})$ where $$p(x)=x^7+x^4+x=x(x^6+x^3+1).$$
Now, it turns out that $2^{2^n}$ is always a root of $q(x)=x^6+x^3+1,$ modulo $19.$
Specifically, since $$q(x^2)=x^{12}+x^6+1=q(x)(x^6-x^3+1),$$
if $x$ is a root of $q$ then $x^2$ is a root of $q.$
So if $x=2^{2^n}$ is a root of $p(x),$ then so is $\left(2^{2^n}\right)^2=2^{2^{n+1}}$ is a root of $p(x).$
So we only need to deal with the case $n=1.$

At heart, what is happening is that the roots of $x^6+x^3+1$ are the primitive 9th roots of $1.$ That is:
$$\frac{x^9-1}{x^3-1}=x^6+x^3+1$$
Squaring a primitive 9th root of $1$ yields another primitive ninth root of $1.$
We can do examples of the same sort, modulo $37,$ such as:
$$37\mid 3^{2^n}+4^{2^n}+ 7^{2^n}$$
Here, $4^4\equiv -3, 4^7\equiv -7\pmod{37}.$
Other cases:
$$73\mid 2^{2^n}+16^{2^n}+18^{2^n}\\
109\mid 4^{2^n}+34^{2^n}+38^{2^n}$$
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