Prove $15\mid 4^{2n}-1$ without induction

Question

Prove without induction that $\forall n \in\mathbb Z$, $15\mid4^{2n}-1$.

$4^{2n} = (4^2)^n = 16^n$.
If $n=1,$ then $(4^2)^n-1=15$, and for $n=2$, it is $255$, which is divisible by $15$.
Using the congruence arithmetic, any $n$ in $4^{2n}-1$ can be expressed as a product of prime factors. Also, there is only one even prime, and any odd prime will be having for any positive integer $k$, an addition of $1+ k.$(even terms).
Say, for $n=5$, with $5=2^2+1$.
And it already proved that any even power of $n$ leads $4^{2n}-1$ to be a multiple of $15.$

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edit Based on the selected answer, have observations, that are possibly implied in that answer, but not explicitly stated.
$4^{2n} -1 = (2^n -1)(2^n+1)(2^{2n}+1) = (2^n -1)(2^n+1)(4^n+1)$
Need consider both even and odd cases separately, for values w.r.t. two modulo $3,5$.

(A) $n$ is even :

(i) check w.r.t. modulo $3$
(a) $ 2^n-1 :: 2\equiv -1 \pmod 3 \implies 2^n \equiv 1 \pmod 3 \implies (2^n-1) \equiv 0 \pmod 3$
(b) $ 2^n+1 :: 2^n +1 \equiv 2 \pmod 3$
(c) $ 4^n+1 :: 4 \equiv 1 \pmod 3 \implies 4^n \equiv 1 \pmod 3$

Finding the product of the three terms w.r.t. modulo $3$, is :
$(\equiv 0 \pmod 3)(\equiv 2 \pmod 3)(\equiv 1 \pmod 3)$
As the first term of the product is a factor of $3$, so divisible by $3$.

(ii) check w.r.t. modulo $5$
(a) $ 2^n-1$ :: Take $n = 2n', 2^2 \equiv -1 \pmod 5\implies$$ 4^{n'} \equiv 1 \pmod 5 $ $\implies (2^n-1) \equiv 0 \pmod 5$
(b) $ 2^n+1$ :: Take $n = 2{n'}, 4^{n'} +1 \equiv 2 \pmod 5$
(c) $ 4^n+1 :: 4 \equiv -1 \pmod 5 \implies 4^n \equiv 1 \pmod 5$

Finding the product of the three terms w.r.t. modulo $5$, is :
$(\equiv 0 \pmod 5)(\equiv 2 \pmod 5)(\equiv 1 \pmod 5)$
As the first term of the product is a factor of $5$, so divisible by $5$.

(B) $n$ is odd :

(i) check w.r.t. modulo $3$
(a) $ 2^n-1 :: 2\equiv -1 \pmod 3 \implies 2^n \equiv -1 \pmod 3 \implies (2^n-1) \equiv -2 \pmod 3$
(b) $ 2^n+1 :: 2^n +1 \equiv 0 \pmod 3$
(c) $ 4^n+1 :: 4 \equiv 1 \pmod 3 \implies 4^n \equiv 1 \pmod 3$

Finding the product of the three terms w.r.t. modulo $3$, is :
$(\equiv -2 \pmod 3)(\equiv 0 \pmod 3)(\equiv 1 \pmod 3)$
As middle term of the product is a factor of $3$, so divisible by $3$.

(ii) check w.r.t. modulo $5$
(a) $ 2^n-1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$
(b) $ 2^n+1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$
(c) $ 4^n+1 :: 4 \equiv -1 \pmod 5 \implies 4^n +1\equiv 0 \pmod 5$

Finding the product of the three terms w.r.t. modulo $5$, is :
$(....)(...)(\equiv 0 \pmod 5)$
As the last term of the product is a factor of $5$, so divisible by $5$.

Don't know how to handle the last case's , sub cases (a), (b) for checking w.r.t. modulo $5$.

Answer 1

note that $$4^{2n}-1=(2^n-1)(2^n+1)(2^{2n}+1)$$ also possible: $$4^{2n}=16^n$$ and $$16\equiv 1 \mod 15$$ so $$16^n\equiv 1^n\equiv 1 \mod 15$$therefore $$16^n-1\equiv 0 \mod 15$$ so no factorization needed

Answer 2

$$4^{2n}-1=16^n-1=(16-1)(16^{n-1}+16^{n-2}+\dots+1)$$

Answer 3

It is $4^{2n}=16^n.$ Note: $$16\equiv 1 \ \ (\mod 3) \Rightarrow 16^n-1\equiv 0 \ \ (\mod 3);$$ $$16\equiv 1 \ \ (\mod 5) \Rightarrow 16^n-1\equiv 0 \ \ (\mod 5).$$ Since it is divisible by both $3$ and $5$, it is also divisible by $15$.

Answer 4

$4 \equiv 1 \mod 3$. So $4^k - 1\equiv 1-1 \equiv 0 \mod 5$ for all $k$. So all $4^k-1$ and all $4^{2n} -1$ are divisible by $3$.

$4 \equiv -1 \mod 5$. So $4^{2n} -1 \equiv (-1^n)^2 - 1 \equiv 1 - 1\equiv 0 \mod 5$ for all even $2n$ so all $4^{2n} -1$ are divisible by $5$.

So $4^{2n} -1$ is divisible by $3*5=15$.