Let

be the set of all pairs

of relatively prime positive integers

with

even and

For

write

where

are positive integers with

odd and define

Prove that

is a function from

to

and that for each

there exists a positive integer

such that

where
If

is a prime number which does not divide

for

prove that the smallest value

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
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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 $\leq x.$