Puzzles
3n+1
Let \(S=\{3n+1:n\in\mathbb{N}\}\) be the set of numbers one more than a multiple of three.
(i) Show that \(S\) is closed under multiplication.
ie. Show that if \(a,b\in S\) then \(a\times b\in S\).
Let \(p\in S\) be irreducible if \(p\not=1\) and the only factors of \(p\) in \(S\) are \(1\) and \(p\). (This is equivalent to the most commonly given definition of prime.)
(ii) Can each number in \(S\) be uniquely factorised into irreducibles?
Show answer & extension
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(i) Let \(a,b\in S\). Then \(\exists \alpha,\beta\in \mathbb{N}\) such that \(a=3\alpha+1\) and \(b=3\beta+1\).
(This says that if \(a\) and \(b\) are in \(S\) then they can be written as a multiple of three plus one.)
$$a\times b=(3\alpha+1)\times (3\beta+1)$$
$$=9\alpha\beta+3\alpha+3\beta+1$$
$$=3(3\alpha\beta+\alpha+\beta)+1$$
This is a multiple of three plus one, so \(a\times b\in S\).
(ii) No, as \(36\times 22=4\times 253\) and 36,22,4 and 253 are all irreducible.
Extension
Try the task again with \(S=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}\).
