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