Using modular arithmetic to help calculate Greatest Common Divisors

In a recent FPM class we discussed how to determine $\mathrm{GCD}(21,2^{70}+5)$.
[See also my earlier post on Greatest Common Divisors]
The answer here was that $21 \mathrel{|} 2^{70}+5$ and so

$\mathrm{GCD}(21,2^{70}+5) = 21\,.$

We discussed various ways to prove this. Some students had shown separately that $3 \mathrel{|} 2^{70}+5$ and $7 \mathrel{|} 2^{70}+5$, from which it follows, e.g. by prime factorization (Fundamental Theorem of Arithmetic, FTA), or by more elementary means, that $21 \mathrel{|} 2^{70}+5$ too.

In the Engagement Session we showed, instead, that $2^{70} \equiv 2^4 = 16$  $(\mathrm{mod}~ 21)$, which gives us

$2^{70}+5 \equiv 16+5 = 21 \equiv 0$  $(\mathrm{mod}~ 21)$

and so $21 \mathrel{|} 2^{70}+5$.
We discovered that $2^{70} \equiv 2^4$  $(\mathrm{mod}~ 21)$ by looking at powers of $2$ mod $21$, and spotting that we had a repeating pattern with period $6$ , so that

$2^{70}=2^{4+66} \equiv 2^4$ $(\mathrm{mod}~ 21)$

We then made this slightly more formal by noting that $2^6=64 = 3 \times 21+1$, so that

$2^6 \equiv 1$ $(\mathrm{mod}~ 21)$,

which explains the repeating pattern. If you want to use the other approach (divisibility by $3$ and by $7)$, I would note that

$2^2=4 \equiv 1$ $(\mathrm{mod}~ 3)$,

and

$2^3=8 \equiv 1$ $(\mathrm{mod}~ 7)$.

Then $2^{70} = (2^2)^{35} \equiv 1^{35}=1$ $(\mathrm{mod}~ 3)$, which gives us

$2^{70}+5 \equiv 1+5 = 6 \equiv 0$ $(\mathrm{mod}~ 3)$,

and so $3 \mathrel{|} 2^{70}+5$ .
[An alternative is to use $2 \equiv -1$ $(\mathrm{mod}~ 3)$, so that $2^{70} \equiv (-1)^{70}=1$ $(\mathrm{mod}~ 3)$, etc.] Similarly $2^{69} =(2^3)^{23} \equiv 1^{23}=1$ $(\mathrm{mod}~ 7)$, which gives us

$2^{70}+5 = 2 \times 2^{69} + 5 \equiv 2 \times 1 + 5 = 7 \equiv 0$ $(\mathrm{mod}~ 7)$

and so $7 \mathrel{|} 2^{70}+5$ .