Last updated: 2019-01-25
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| File | Version | Author | Date | Message |
|---|---|---|---|---|
| Rmd | d5bcc6f | John Blischak | 2019-01-25 | Add style and navigation. |
| html | 0520a16 | John Blischak | 2019-01-25 | Build site. |
| Rmd | 1c3d6a1 | John Blischak | 2019-01-25 | Add some proofs from discrete math. |
If \(k\) is any odd integer and \(m\) is any even integer, then \(k^{2} + m^{2}\) is odd.
Proof: Let \(k\) and \(m\) be any integers, and suppose that \(k\) is odd and \(m\) is even. By definition of odd, \(k = 2r + 1\) for some integer \(r\). By definition of even, \(m = 2s\) for some integer \(s\). It follows that
\[ k^{2} + m^{2} = (2r + 1)^{2} + (2s)^{2} \]
\[ = 2(2r^{2} + 2r + 2s^{2}) + 1. \]
\(2r^{2} + 2r + 2s^{2}\) is an integer because sums and products of integers are integers. By definition of odd, \(k^{2} + m^{2}\) is odd. \(\square\)
For all integers \(a\), \(b\), and \(c\), if \(a|b\) and \(a|c\) then \(a|(2b - 3c)\).
Proof: Let \(a\), \(b\), and \(c\) be any integers and suppose \(a|b\) and \(a|c\). By definition of divisibilty, \(b = ra\) and \(c = sa\) for some integers \(r\) and \(s\). It follows that
\[ 2b - 3c = 2(ra) - 3(sa) = a(2r - 3s). \]
\(2r - 3s\) is an integer because sums and products of integers are integers. Thus by definition of divisibility, \(a|(2b - 3c)\). \(\square\)
The square of any integer has the form \(3k\) or \(3k + 1\) for some integer \(k\).
Proof: Let \(n\) be any integer, then by QRT (with \(d = 3\)) \(n = 3q + r\) for some integers \(q\) and \(r\) with \(0 \leq r < 3\).
Case 1: Suppose \(r = 0\). Then \(n = 3q + 0 = 3q\) and \(n^{2} = (3q)^{2} = 3(3q^{2})\). Let \(k = 3q^{2}\), which is an integer because products of integers are integers. Thus \(n^{2} = 3k\).
Case 2: Suppose \(r = 1\). Then \(n = 3q + 1\) and \(n^{2} = (3q + 1)^{2} = 3(3q^{2} + 2q) + 1\). Let \(k = 3q^{2} + 2q\), which is an integer because sums and products of integers are integers. Thus \(n^{2} = 3k + 1\).
Case 3: Suppose \(r = 2\). Then \(n = 3q + 2\) and \(n^{2} = (3q + 2)^{2} = 3(3q^{2} + 4q + 1) + 1\). Let \(k = 3q^{2} + 4q + 1\), which is an integer because sums and products of integers are integers. Thus \(n^{2} = 3k + 1\).
Therefore, in all cases \(n^{2} = 3k\) or \(n^{2} = 3k + 1\). \(\square\)