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Solve $N(U_f + {0 \over 1} ) = N({1 \over 0})$ = unknot, \hfil \break
$N(U_f + {t \over w} ) = N({0 \over 1})$ = unlink of two components.


Method 1:  

1.)  Since $N({0 \over 1})$ is a $(2, p)$ torus link, $ N({1 \over 0})$ = unknot, and $E$ is rational, $U_f$ is rational.  

Suppose $U_f = {a \over b}$.  Then 

 $N(U_f + {0 \over 1} ) = N({a \over b} + {0 \over 1} ) = N({a \over b}) = N({1 \over 0})$.  

Hence $a = 1$  since we can take $a$ to be nonnegative (${a \over b} = { -a \over -b}$).   

Thus $U_f = {1 \over b}$ .

$N(U_f + {t \over w} ) = N({1 \over b} + {t \over w} ) = N({w + tb \over t}) =  N({0 \over 1})$

Hence $w + tb = 0$.  Thus $w = -tb$.  Since $gcd(w, t) = 1$ and we can take $t$ to be nonnegative, t = 1.  Hence $w = -b$.

Thus the solutions to the above system of tangle equations is $U_f = {1 \over b}$ and ${t \over w} = -{1 \over b}$  where $b$ is an arbitrary integer.


Method 3:  Use KnotPlot 

\eject
Method 2:  

Solve $N(U_f + {0 \over 1} ) = N({1 \over 0}) = N({a \over b})$ = unknot, \hfil \break
$N(U_f + {t \over w} ) = N({0 \over 1}) = N({z \over v})$ = 2 component unlink.

$a = 1, b = 0, x = 0, y = -1, z = 0, v = 1, $ \hfil \break and $v'$ is any integer such that $v'v^{\pm 1} = 1$mod $z$

By corollary 2 if $w \not\cong \pm 1 $ mod $t$ or if $U_f$ is rational, then ${t \over w} = {xz - av' \over bv' - yz - kt} = {v' \over -kt} =$ \hfil \break
and $U = {a \over b + ka} = {1 \over k}$ \hfil \break
Or ${t \over w} = {bz - av' \over xv' - yz - kt} = {v' \over -kt} =$ \hfil \break
and $U = {a \over x + ka} = {1 \over k}$

By Thm 3, if $w \cong \pm 1$ mod $t$, $N({z \over v}) = N({tp(pb - qa) \pm a \over tq(pb - qa) \pm b})$

Hence $ tp(pb - qa) \pm a = z $ or $-z$.  \hfil \break
Thus $ tp(pb - qa) = z \mp a $ or $-z \mp a$.    

Hence $tp | (z \mp a)$.  

IF $t \not= \pm 1$, then $U = ({da - jb \over pb - qa} + {j \over p}) \circ (h, 0)$ or $({j \over p} +{da - jb \over pb - qa} )\circ (h, 0)$. Hence if $|z \mp a|$ is prime, $U$ is rational. 


If $t = \pm 1$, then $U$ is rational by Hirasawa and Shimokawa.


\eject

Solve $N(U_f' + {f_1 \over g_1} ) = N({1 \over 0}) = N({a \over b})$ = unknot, \hfil \break
$N(U_f' + {f_2 \over g_2} ) = N({0 \over 1}) = N({z \over v})$ = 2 component unlink.

given that $U_f = -{1 \over k}$ and ${t \over w} = {1 \over k}$ are the solutions to 
   $N(U_f + {0 \over 1} ) = N({1 \over 0}) = N({a \over b})$ = unknot, \hfil \break
$N(U_f + {t \over w} ) = N({0 \over 1}) = N({z \over v})$ = 2 component unlink.

Suppose ${f_1 \over g_1} = (c_1, ..., c_n)$, $f_1 = E[c_1, ..., c_n],$
$ g_1 = E[c_1, ..., c_{n-1}], e_1 = E[c_2, ..., c_n], i_1 = E[c_2, ..., c_{n-1}]$.


Then ${f_2 \over g_2} = { te_1 + wf_1 \over ti_1 + wg_1} = (b_1, ..., b_k + c_1, ..., c_n)$ where ${t \over w} = (b_1, ..., b_k)$.

Since ${t \over w} = {1 \over k} = (k, 0)$, \hfil \break
${f_2 \over g_2} = { e_1 + kf_1 \over i_1 + kg_1} = ( k, 0 + c_1, ..., c_n) = ( k, c_1, ..., c_n)$.

$U_f' = U_f \circ( -c_1, ..., -c_n) = (-k, 0) \circ ( -c_1, ..., -c_n) = (-k, 0 + -c_1, ..., -c_n)= (-k,  -c_1, ..., -c_n)  $

If $U_f = {a \over b'}$, then $U_f' = U_f \circ( -c_1, ..., -c_n) = {-f_1 b' + e_1a \over g_1b' - i_1a}$.

Since $U_f = {1 \over -k}$, $U_f' = {f_1k + e_1 \over g_1 k - i_1}$

Note: (1) = (0, 1, 1), but 

\centerline{$(3) = (2) \circ (1) \not= (2) \circ (0, 1, 1) = (2, 1, 1)$}

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