From Solving unoriented tangle equations involving 4-plats Theorem 3 N(U + \frac{0}{1}) = N(\frac{a}{b}) and N(U + \frac{t}{w}) = N(\frac{z}{v}) where N(\frac{a}{b}) and N(\frac{z}{v}) are unoriented 4-plats and U is a generalized M-tangle if and only if the following hold:

(a) If w ≅ ±̸1 mod t, then there exists an integer, b′ such that b′b^{±1} = 1 mod a, and for any integers x and y such that b′x − ay = 1,

N(\frac{z}{v} ) = N(\frac{tb′ + wa}{ty + wx})

In this case, U = \frac{a}{b′} for all b′ satisfying the above.

(b) If w ≅ ±1 mod t, then there exists relatively prime integers, p and q, where p may be chosen to be positive, such that

N(\frac{z}{v} ) = N(\frac{tp(pb − qa) ± a}{tq(pb − qa) ± b})

In this case, the solutions for U are (\frac{da−jb}{pb−qa} + \frac{j}{p})∘(h, 0) and (\frac{j}{p} + \frac{da−jb}{pb−qa})∘(h, 0), for all p, q satisfying the above, d and j are any integers such that pd−qj = 1, and h = \frac{−w±1}{t} where the ± sign agrees with that in (e-w1) (note, the choice of j and d such that pd − qj = 1 has no effect on U). Corollary 1 If |t| > 1 or if N(\frac{a}{b}) = N(\frac{1}{0}), N(\frac{z}{v}) = N(\frac{2k}{1} ), then theorem t-un gives all solutions to the system of tangle equations: N(U + \frac{0}{1}) = N(\frac{a}{b}) and N(U + \frac{t}{w}) = N(\frac{z}{v}) where N(\frac{a}{b}) and N(\frac{z}{v}) are unoriented 4-plats. Corollary 2 Suppose bx − ay = 1, N(U + \frac{0}{1}) = N(\frac{a}{b}) and N(U + \frac{t}{w}) = N(\frac{z}{v}) where N(\frac{a}{b}) and N(\frac{z}{v}) are unoriented 4-plats. If w ≅ ±̸1 or if U is rational, then \frac{t}{w} = \frac{xz−av′}{bv′−yz−kt} and U = \frac{a}{b+ka} OR \frac{t}{w} = \frac{bz−av′}{xv′−yz−kt} and U = \frac{a}{x+ka} where v′ is any integer such that v′v^{±1} = 1 mod z. If w ≅ ±1 mod t, then t divides z ∓ a. Theorem 4 A (0, \frac{t}{w}) move is equivalent to a (0, \frac{c}{d}) move if and only if \frac{c}{d} = \frac{t}{w−ht} for some h. Theorem 5 Suppose f_{1}/g_{1} = (c_{1}, ..., c_{n}), n odd, where f_{1} = E[c_{1}, ..., c_{n}] and g_{1} = E[c_{1}, ..., c_{n−1}]. Let e_{1} = E[c_{2}, ..., c_{n}], i_{1} = E[c_{2}, ..., c_{n−1}]. If \frac{f_{2}}{g_{2}} = (d_{1}, ..., d_{m}), let \frac{t}{w} = \frac{g_{1}f_{2}−g_{2}f_{1}}{e_{1}g_{2}−i_{1}f_{2}} = (d_{1}, ..., d_{m} − c_{n}, −c_{n−1}, ..., −c_{1}) and U′ = U ∘ (c_{n}, ..., c_{1}) (or equivalently, if \frac{t}{w} = (b_{1}, ..., b_{k}), let \frac{f_{2}}{g_{2}} = \frac{te_{1}+wf_{1}}{ti_{1}+wg_{1}} = (b_{1}, ..., b_{k} + c_{1}, ..., c_{n}), and U = U′ ∘ (−c_{1}, ..., −c_{n})), then for any pair of knots K_{1}, K_{2},

N(U + \frac{f_{1}}{g_{1}} ) = K_{1} N(U + \frac{f_{2}}{g_{2}} ) = K_{2}

if and only if

N(U′ + \frac{0}{1}) = K_{1} N(U′ + \frac{t}{w}) = K_{2}

Hence an (\frac{f_{1}}{g_{1}} , \frac{f_{2}}{g_{2}} ) move is equivalent to a (0, \frac{g_{1}f_{2}−g_{2}f_{1}}{e_{1}g_{2}−i_{1}f_{2}} ) move, and similarly, a (0, \frac{t}{w}) move is equivalent to an (\frac{f_{1}}{g_{1}} , \frac{te_{1}+wf_{1}}{ti_{1}+wg_{1}} ).

Moreover if (\frac{f_{1}}{g_{1}} , \frac{f_{2}}{g_{2}} ) is equivalent to (0, \frac{t}{w}) then there exists e_{1} and i_{1} such that g_{1}e_{1} − f_{1}i_{1} = 1 an\frac{g_{1}f_{2}−g_{2}f_{1}}{e_{1}g_{2}−i_{1}f_{2}} (or equivalently, \frac{f_{2}}{g_{2}} = \frac{te_{1}+wf_{1}}{ti_{1}+wg_{1}} ).

Theorem 6 (B = \frac{f_{1}}{g_{1}} )

Suppose \frac{f_{1}}{g_{1}} = (c_{1}, ..., c_{n}), n odd, 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}], t = g_{1}f_{2} − g_{2}f_{1}, w = e_{1}g_{2} − i_{1}f_{2}. N(U + \frac{f_{1}}{g_{1}} ) = N(\frac{a}{b}) and N(U + \frac{f_{2}}{g_{2}} ) = N(\frac{z}{v}) where N(\frac{a}{b}) and N(\frac{z}{v}) are unoriented 4-plats and U is a generalized M-tangle if and only if the following hold:

(a) If w ≅ ±̸1 mod t, then there exists an integer, b′ such that b′b^{±1} = 1 mod a, and for any integers x and y such that b′x − ay = 1,

N(\frac{z}{v} ) = N(\frac{tb′ + wa}{ty + wx})

In this case, U = \frac{a}{b′} ∘ (−c_{1}, ..., −c_{n}) = \frac{−f_{1}b′+e_{1}a}{g_{1}b′−i_{1}a} for all b′ satisfying the above.

(b) If w ≅ ±1 mod t, then there exists relatively prime integers, p and q, where p may be chosen to be positive, such that

N(\frac{z}{v} ) = N(\frac{tp(pb − qa) ± a}{tq(pb − qa) ± b})

In this case, U = (U_{1} +U_{2})∘(h, −c_{1}, ..., −c_{n})) and U = (U_{2} +U_{1})∘(h, −c_{1}, ..., −c_{n}) where U_{1} = \frac{j}{p}, U_{2} = \frac{da−jb}{pb−qa} are both solutions for U, for all p, q satisfying the above, and d and j are any integers such that pd − qj = 1, and h = \frac{−w±1}{t} (note, the choice of j such that pd − qj = 1 has no effect on U).

If |f_{1}g_{2} − f_{2}g_{1}| > 1 or if |f_{1}g_{2} − f_{2}g_{1}| = 1, N(\frac{a}{b}) = N(\frac{1}{0}) and N(\frac{z}{v}) = N(\frac{2k}{1} ), then the above list of solutions to the system of equations, N(U + \frac{f_{1}}{g_{1}} ) = N(\frac{a}{b}) and N(U + \frac{f_{2}}{g_{2}} ) = N(\frac{z}{v}), is complete.