Using induction, I have to prove that for any $n \in \mathbb{N}^*$ the following is true:
$$2^{2^n} + 3^{2^n} + 5^{2^n} \equiv 0 \pmod{19}$$
I've never done modular arithmetic before, but after reading online a bit I understood that the above notation means that when we divide the $2$ terms to the left and to the right of the $\equiv$ symbol, we get the same remainder. So that means that since the right term is $0$ $\Rightarrow$ the left term is a multiple of $19$. So we have:
$$2^{2^n} + 3^{2^n} + 5^{2^n} = 19a$$
with $a \in \mathbb{N}^*$. The base case is obvious:
$$p(1): 2^2 + 3^2 + 5 ^ 2 = 38 = 2 \cdot 19$$
so $p(1)$ is evidently true. I have problems with the induction step. Let's assume $p(k)$ true, so we have:
$$p(k):2^{2^k} + 3^{2^k} + 5^{2^k} = 19a$$
for some $a \in \mathbb{N}^*$. So now we have to show that:
$$p(k+1) : 2^{2^{k+1}} + 3^{2^{k+1}} + 5^{2^{k+1}}$$
is also a multiple of $19$. I have problems with this. I tried squaring what I got at $p(k)$ since that would get me the terms that I want:
$$\hspace{6.7cm} 2^{2^k} + 3^{2^k} + 5^{2^k} = 19a \hspace{6cm} ()^2$$
$$2^{2^{k+1}} + 3^{2^{k+1}} + 5^{2^{k+1}} + 2 \cdot 2^{2^k} \cdot 3^{2^k} + 2 \cdot 2^{2^k} \cdot 5^{2^k} + 2 \cdot 3^{2^k} \cdot 5^{2^k} = 361a^2$$
$$2^{2^{k+1}} + 3^{2^{k+1}} + 5^{2^{k+1}} + 2 (6^{2^k} + 10^{2^k} + 15^{2^k}) = 361 a^2$$
$$2^{2^{k+1}} + 3^{2^{k+1}} + 5^{2^{k+1}} = 361 a^2 - 2 (6^{2^k} + 10^{2^k} + 15^{2^k})$$
Since $361$ is a multiple of $19$, in order to show that the left-hand side is a multiple of $19$ I would have to show that $6^{2^k} + 10^{2^k} + 15^{2^k}$ is also a multiple of $19$, so I just exchanged a problem I couldn't solve for another problem I can't solve.
How should I approach this?