32 CHAPTER 2 STRATEGIES FOR INVESTIGATING PROBLEMS
The pattern seems simple: f(m) increases by 1 as m increases, until m is a perfect
square , in which case f(m) increases by 2 (boldface). Whenever you observe a pattern,
you should try to see why it is true. In this case, it is not hard to see what is going on.For example, if 9 :::; m < 16, the quantity l vrn J has the constant value 3. Hence
f(m) = m+3 for 9 :::; m < 16. Likewise, f(m) = m+4 for^16 :::; m < 25, which
accounts for the "skip" that happens at perfect squares.Now that we understand f pretty well, it is time to look at repeated iterations of
this function. Again, experiment and make a table! We will use the notationr(m) = f(f(-·· f(m)·· .)), (2)
where the right-hand side contains r fs, and we will indicate squares with boldface.
Notice that we don't bother trying out values of m that are squares, since then we
would trivially be done. Instead, we start with values of m that are one more than a
perfect square, etc. Also, even when we achieve a perfect square, we continue to fill in
the table. This an important work habit:Don't skimp on experimentation! Keep messing around until you think
you understand what is going on. Then mess around some more.
m f(m) F(m) p(m) f 4 (m) p(m) r(m)
5 7 9 12 15 18 22
6 8 10 13 16 20 24
7 9 12 15 18 22 26
8 10 13 16 20 24 28
50 57 64 72 80 88 97
51 58 65 73 81 90 99
101 III 121 132 143 154 166
102 112 122 133 144 156 168
103 113 123 134 145 157 169
Now more patterns emerge. It seems that if m has the form n^2 + 1, where n is an
integer, then f^2 (m) is a perfect square. Likewise, if m has the form n^2 + 2, then r (m)
appears to be a perfect square. For numbers of the form n^2 + 3, such as 7 and 103, it is
less clear: f^6 (103) is a perfect square , yet f(7) is a perfect square while f^6 (7) is not.
So we know that the following "conjecture" is not quite correct:lfm = n^2 +b, then f^2 b(m) is a perfect square.
Nevertheless, the wishful thinking strategy demands that we at least examine thisstatement. After all, we wish to prove that for any m, there will be an r such that r (m)
is a perfect square. Before we dive into this, let's pause and consider: what is the chief
difficulty with this problem? It is the perplexing l vrnJ term. So we should first focuson this expression. Once we understand it, we will really understand how f(m) works.
Define g(m) = l vrnJ. Then what is g(n^2 + b) equal to? If b is "small enough," then
g(n^2 + b) = n. We can easily make this more precise, either by experimenting or just
thinking clearly about the algebra. For what values of m is g(m) = n? The answer:
n^2 :::; m < (n + 1 f = n^2 + 2n + 1.