Let
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
be the set of all pairs
![(m,n)](/media/m/4/2/0/4202026f5dcd727e559bf78936cde057.png)
of relatively prime positive integers
![m,n](/media/m/5/0/b/50b28d5a8b3362b2ac435400d318b3af.png)
with
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
even and
![m < n.](/media/m/1/c/6/1c6e4741c90588a09273f66b63520bef.png)
For
![s = (m,n) \in S](/media/m/d/3/e/d3eb307d061e21397f7b82f7ae51c00c.png)
write
![n = 2^k \cdot n_o](/media/m/f/f/f/fff17499e5fd54825a5d6b3eb3c3d559.png)
where
![k, n_0](/media/m/7/0/4/7041c39488c2e9f377b5e815d4c645ec.png)
are positive integers with
![n_0](/media/m/2/2/a/22a23a21b7b74b6120b58e119a1f870c.png)
odd and define
![f(s) = (n_0, m + n - n_0).](/media/m/8/9/b/89b44ccca42a5cf3dcd7c1cc923a6675.png)
Prove that
![f](/media/m/9/9/8/99891073047c7d6941fc8c6a39a75cf2.png)
is a function from
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
to
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
and that for each
![s = (m,n) \in S,](/media/m/d/0/7/d071ea3787b022739f9734c90e0143a3.png)
there exists a positive integer
![t \leq \frac{m+n+1}{4}](/media/m/d/3/9/d39fa227cc1a5dbf90075baf9cc5974c.png)
such that
![f^t(s) = s,](/media/m/8/e/e/8ee6af68413890a69086e74d76b6cff6.png)
where
If
![m+n](/media/m/1/2/c/12cfeebf074af065b2efa21bce4eb0fe.png)
is a prime number which does not divide
![2^k - 1](/media/m/6/8/c/68c4686c193162b38f65e49c5734152d.png)
for
![k = 1,2, \ldots, m+n-2,](/media/m/7/c/3/7c3ed40df0962e98ca615760b83cc606.png)
prove that the smallest value
![t](/media/m/7/f/6/7f630d3904cfcd77d22bd7938423df6c.png)
which satisfies the above conditions is
![\left [\frac{m+n+1}{4} \right ]](/media/m/8/7/8/878d7261cbd40fe8c838d36d2b94fea7.png)
where
![\left[ x \right]](/media/m/8/4/7/847a3b7449538c2b99179a2953e7f9e0.png)
denotes the greatest integer
%V0
Let $S$ be the set of all pairs $(m,n)$ of relatively prime positive integers $m,n$ with $n$ even and $m < n.$ For $s = (m,n) \in S$ write $n = 2^k \cdot n_o$ where $k, n_0$ are positive integers with $n_0$ odd and define $$f(s) = (n_0, m + n - n_0).$$ Prove that $f$ is a function from $S$ to $S$ and that for each $s = (m,n) \in S,$ there exists a positive integer $t \leq \frac{m+n+1}{4}$ such that $$f^t(s) = s,$$ where $$f^t(s) = \underbrace{ (f \circ f \circ \cdots \circ f) }_{t \text{ times}}(s).$$
If $m+n$ is a prime number which does not divide $2^k - 1$ for $k = 1,2, \ldots, m+n-2,$ prove that the smallest value $t$ which satisfies the above conditions is $\left [\frac{m+n+1}{4} \right ]$ where $\left[ x \right]$ denotes the greatest integer $\leq x.$