divisibility - Prove by mathematical induction that $forall n in mathbb{N}: 20~|~4^{2n} + 4$

Prove by mathematical induction that $\forall n \in \mathbb{N}: 20~|~4^{2n} + 4$

Step 1: Show that the statement is true for n = 1:

$4^{2 \cdot 1} + 4 = 20$

Since $20~|~20$, the base case is completed.

Step 2: Show that if the statement is true for n = p, it is true for n = p + 1:

The general idea I had here was to try to get $4^{2(p+1)} + 4$ to look like the original statement, perhaps multiplied with a factor or a sum where each group of terms were divisible by 20.

$4^{2(p+1)} + 4 = 4^{2p+2} + 4$

By one of the exponential rules, we have:

$4^{2p+2} + 4 = 4^{2p} \cdot 4^{2} + 4$

At this point, we could use one of the 16 powers of 4 and together with the 4 term and show that this was divisible by 20, but there does not seem to be any clear reason why $20~|~4^{2p} \cdot 15$. The number $4^{2p}$ always have an even exponent and $4 \cdot 5 = 20$, but not sure how to proceed from here. Any suggestions?

Answer

Try

$$4^{2p+2}+4=16(4^{2p} +4)-60.$$

Remark: From a number-theoretic point of view, an approach like the one of BRIC-Fan is more natural. It is enough to prove that $4^{2n-1}+1\equiv 0\pmod{5}$. Note that $4\equiv -1\pmod{5}$. So we want to show that $(-1)^{2n-1}+1\equiv 0\pmod{5}$. In principle one proves this by induction. But since $-1$ to an odd power is $-1$, one can view the result as obvious.