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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 \(n^{2} = 3k + 1\). \(\square\)
The following proof is from Discrete Mathemtics with Applications, 4th Edition by Susanna S. Epp (p. 407).
For any positive real numbers \(b\), \(c\), and \(x\), with \(b \neq 1\) and \(c \neq 1\),
\[ log_{c}x = \frac{log_{b}x}{log_{b}c}.\]
Proof: Suppose positive real numbers \(b\), \(c\), and \(x\), with \(b \neq 1\) and \(c \neq 1\). Let
\[ \begin{align*} u = log_{b}c \>\>\>\> (1) \\ v = log_{c}x \>\>\>\> (2) \\ w = log_{b}x \>\>\>\> (3) \end{align*} \]
Then, by definition of logarithm,
\[ \begin{align*} c = b^u \>\>\>\> (1') \\ x = c^v \>\>\>\> (2') \\ x = b^w \>\>\>\> (3') \end{align*} \] Substituting (1’) into (2’) and using one of the laws of exponents, \((b^{u})^v = b^{uv}\), gives
\[ x = c^v = (b^u)^v = b^{uv} \]
But by (3’), \(x = b^w\) also. Hence
\[ b^{uv} = b^{w}, \]
and so by the one-to-oneness of the exponential function (if \(b^u = b^v\) then \(u = v\)),
\[ uv = w. \]
Substituting from (1), (2), and (3) gives that
\[ (log_{b}c)(log_{c}x) = log_{b}x. \]
And dividing both sides by \(log_{b}c\) (which is nonzero because \(c \neq 1\)) results in
\[ log_{c}x = \frac{log_{b}x}{log_{b}c}. \>\>\>\> \square\]