Suppose:
Inspection of the rule for multiplication will reveal that the factor n must also appear in every component of A.
Were B and C reciprocals (i.e. multiplicative inverses), A would by definition equal 1:
⟨ B_{q}, B_{r}, B_{s}, B_{t} ⟩ × ⟨ C_{q}, C_{r}, C_{s}, C_{t} ⟩ = A = ⟨ 0, +1, 0, −1 ⟩
However, ⟨ 0, +1, 0, −1 ⟩ clearly cannot contain that factor n. Thus in the case where a quintright has a reciprocal, its four components cannot have a common nontrivial factor. (Two or three of the components might share a nontrivial common factor, but not all four.)
Recalling the division problem C = A ÷ B from the main page, write A = 1 = ⟨ 0, +1, 0, −1 ⟩ as a step toward obtaining C as the reciprocal of B:
0  = C_{q}(B_{r})  + C_{r}(B_{q} + B_{s})  + C_{s}(B_{r} + B_{t})  + C_{t}(B_{s}) 
+1  = C_{q}(3B_{q} + B_{s})  + C_{r}(2B_{r} + B_{t})  + C_{s}(B_{q} + 2B_{s})  + C_{t}(B_{r} + B_{t}) 
0  = C_{q}(B_{r} + B_{t})  + C_{r}(B_{q})  + C_{s}(− B_{t})  + C_{t}(B_{q} − B_{s}) 
−1  = C_{q}(B_{s} − 2B_{q})  + C_{r}(− B_{r} − B_{t})  + C_{s}(B_{q} − 3B_{s})  + C_{t}(− B_{r} − 2B_{t}) 
As before, regard this as a system of four linear equations in the four unknowns C_{q}, C_{r}, C_{s}, C_{t} and solve; if the four results are integers, division was successful. A standard application of Cramer's rule will give the solutions. For convenience, give names to the determinants of some matrices:
d_{u} = det 

d_{q} = det 

d_{r} = det 

d_{s} = det 

d_{t} = det 

Then the quintright C = 1 ÷ B = ⟨ d_{q} ÷ d_{u}, d_{r} ÷ d_{u}, d_{s} ÷ d_{u}, d_{t} ÷ d_{u} ⟩ exists whenever d_{u} ≠ 0 and each of the four component division operations yields no remainder. If d_{u} is ±1, the four divisions will patently be without remainder; but the converse of this is open.
In fact, −1 can be rejected as a possibility for the value of d_{u}. Expressed as a polynomial, d_{u} expands to an eighteenterm quartic form, in other words a polynomial where every term is of exactly the same degree:
+ 5B_{q}^{4} + 1B_{r}^{4} + 5B_{s}^{4} + 1B_{t}^{4} 
− 10B_{q}^{3}B_{s} + 6B_{r}^{3}B_{t} + 10B_{q}B_{s}^{3} + 6B_{r}B_{t}^{3} 
− 5B_{q}^{2}B_{r}^{2} − 5B_{q}^{2}B_{s}^{2} − 10B_{q}^{2}B_{t}^{2} − 10B_{r}^{2}B_{s}^{2} + 11B_{r}^{2}B_{t}^{2} − 5B_{s}^{2}B_{t}^{2} 
− 10B_{q}^{2}B_{r}B_{t} + 10B_{q}B_{r}^{2}B_{s} − 10B_{r}B_{s}^{2}B_{t} − 10B_{q}B_{s}B_{t}^{2} 
Now consider d_{u} in modulo 5, as obtained by discarding obvious multiples of 5:
1B_{r}^{4} + 6B_{r}^{3}B_{t} + 11B_{r}^{2}B_{t}^{2} + 6B_{r}B_{t}^{3} + 1B_{t}^{4}
Subtract 10B_{r}^{3}B_{t} + 5B_{r}^{2}B_{t}^{2} + 10B_{r}B_{t}^{3} (which is a multiple of five) in order to yield:
1B_{r}^{4} − 4B_{r}^{3}B_{t} + 6B_{r}^{2}B_{t}^{2} − 4B_{r}B_{t}^{3} + 1B_{t}^{4}
which factors into
(B_{r} − B_{t})^{4}
which for any integer (B_{r} − B_{t}) equals 0 or 1 in modulo 5. Hence d_{u} cannot equal −1.
Something can also be said about the behavior of d_{u} in modulo 4. Each of B_{q}, B_{r}, B_{s}, and B_{t} is even or odd, making 16 = 2^{4} combinations, and each of these is easy to investigate separately. It turns out that in modulo 4, d_{u} always equals 0 or 1, as detailed in the following table:
B_{q}  B_{r}  B_{s}  B_{t}  d_{u} in modulo 4 

even  even  even  even  0 
even  odd  odd  even  0 
odd  even  even  odd  0 
odd  odd  odd  odd  0 
twelve other combinations  1 
If d_{u} must equal either 0 or 1 in each of modulos 4 and 5, then d_{u} must equal 0, 1, 5, or 16 in modulo 20.
Noteworthy is that d_{u} is preserved under multiplication. Consider any two quintrights A and B. Let the symbol d_{u}(B) represent d_{u} as calculated on B_{q}, B_{r}, B_{s}, and B_{t}, exactly as above. Symbol d_{u}(A) works the same way. Then d_{u}(AB) = d_{u}(A) d_{u}(B).
If a nonzero quintright does not have a reciprocal, it still has something that is almost a reciprocal. For any nonzero quintright A, define the governor as the smallest positive integer n such that the quotient n ÷ A = ⟨ 0, +n, 0, −n ⟩ ÷ A does exist. In the four examples of the following table, each B = n ÷ A is the almostreciprocal of its respective A:
A  gov (A)  B  gov (B)  AB 

⟨ +1, +4, +7, +5 ⟩ ⟨ −1, +4, +7, +6 ⟩ ⟨ +1, +2, +7, +6 ⟩ ⟨ +1, +4, +7, +6 ⟩  401 3919 1501 1  ⟨ +106, −215, +137, +9 ⟩ ⟨ −365, +668, +259, −734 ⟩ ⟨ −49, +26, +317, −378 ⟩ ⟨ −1, +4, −7, +6 ⟩  401 3919 1501 1  ⟨ 0, +401, 0, −401 ⟩ = 401 ⟨ 0, +3919, 0, −3919 ⟩ = 3919 ⟨ 0, +1501, 0, −1501 ⟩ = 1501 ⟨ 0, +1, 0, −1 ⟩ = 1 
Conjecture. The governor can be calculated using the greatest common factor (gcf) and least common multiple (lcm) functions:
gcf (d_{q}, d_{r}, d_{s}, d_{t}) × lcm (d_{q}, d_{r}, d_{s}, d_{t}, d_{u}) ÷ lcm (d_{q}, d_{r}, d_{s}, d_{t})
This calculation is apt to cause integer overflow on computers.
Given a quintright that has a reciprocal, we can find others. These transformations are verifiable mechanically:
If ⟨ A_{q}, A_{r}, A_{s}, A_{t} ⟩ × ⟨ B_{q}, B_{r}, B_{s}, B_{t} ⟩ = 1, then
If C and D have reciprocals, so does CD. In fact, the quintrights that have reciprocals form a group under the multiplication operation. The group is infinite; whether it has finitely many generators is an open question.
The following is a list of selected reciprocal pairs. A decimal approximation has been written within the angle brackets.
⟨ 0, 0, 0, +1 ≈ 0.618034 ⟩ × ⟨ 0, +1, 0, 0 ≈ 1.618034 ⟩ = 1 ⟨ 0, 0, +1, 1 ≈ 0.557537 ⟩ × ⟨ 0, 0, +1, +1 ≈ 1.793604 ⟩ = 1 ⟨ 0, 0, +1, +2 ≈ 2.411638 ⟩ × ⟨ 3, +6, 2, 2 ≈ 0.414656 ⟩ = 1 ⟨ 0, +1, 0, 2 ≈ 0.381966 ⟩ × ⟨ 0, +2, 0, 1 ≈ 2.618034 ⟩ = 1 ⟨ 0, +1, +1, 1 ≈ 2.175571 ⟩ × ⟨ +1, 2, +1, +1 ≈ 0.459650 ⟩ = 1 ⟨ 0, +1, +2, +2 ≈ 5.205243 ⟩ × ⟨ +2, 2, 4, +7 ≈ 0.192114 ⟩ = 1 ⟨ 0, +1, +3, +3 ≈ 6.998847 ⟩ × ⟨ +3, 6, +3, +1 ≈ 0.142881 ⟩ = 1 ⟨ 0, +1, +4, +5 ≈ 9.410486 ⟩ × ⟨ 32, +61, 20, 23 ≈ 0.106264 ⟩ = 1 ⟨ 0, +2, 0, 5 ≈ 0.145898 ⟩ × ⟨ 0, +5, 0, 2 ≈ 6.854102 ⟩ = 1 ⟨ 0, +2, +1, 2 ≈ 3.175571 ⟩ × ⟨ +1, +2, 2, 4 ≈ 0.314904 ⟩ = 1 ⟨ 0, +3, 0, 1 ≈ 4.236068 ⟩ × ⟨ 0, 1, 0, +3 ≈ 0.236068 ⟩ = 1 ⟨ 0, +3, +2, 4 ≈ 4.733107 ⟩ × ⟨ 6, +12, 4, 5 ≈ 0.211278 ⟩ = 1 ⟨ 0, +4, +3, 5 ≈ 6.908678 ⟩ × ⟨ 0, 4, +3, +5 ≈ 0.144746 ⟩ = 1 ⟨ 0, +6, +4, 7 ≈ 10.084248 ⟩ × ⟨ +12, +15, 20, 38 ≈ 0.099165 ⟩ = 1 ⟨ +1, +1, 0, 0 ≈ 3.520147 ⟩ × ⟨ +1, 1, 0, 0 ≈ 0.284079 ⟩ = 1 ⟨ +1, +1, 0, 1 ≈ 2.902113 ⟩ × ⟨ +1, 1, 1, +2 ≈ 0.344577 ⟩ = 1 ⟨ +1, +1, +1, 0 ≈ 4.695718 ⟩ × ⟨ 1, +1, +2, 3 ≈ 0.212960 ⟩ = 1 ⟨ +1, +1, +2, 2 ≈ 4.635220 ⟩ × ⟨ 11, 13, +18, +34 ≈ 0.215739 ⟩ = 1 ⟨ +1, +2, 0, 0 ≈ 5.138181 ⟩ × ⟨ 2, 2, +3, +6 ≈ 0.194621 ⟩ = 1 ⟨ +1, +2, 0, 2 ≈ 3.902113 ⟩ × ⟨ 2, +4, 1, 2 ≈ 0.256271 ⟩ = 1 ⟨ +1, +2, +1, 0 ≈ 6.313752 ⟩ × ⟨ 1, +2, 1, 0 ≈ 0.158384 ⟩ = 1 ⟨ +1, +2, +1, 1 ≈ 5.695718 ⟩ × ⟨ 0, 1, +1, +1 ≈ 0.175571 ⟩ = 1 ⟨ +1, +3, +2, 1 ≈ 8.489322 ⟩ × ⟨ +11, 21, +7, +8 ≈ 0.117795 ⟩ = 1 ⟨ +1, +3, +5, +4 ≈ 15.106203 ⟩ × ⟨ 13, +15, +22, 41 ≈ 0.066198 ⟩ = 1 ⟨ +1, +4, +1, 2 ≈ 8.313752 ⟩ × ⟨ +3, +4, 5, 10 ≈ 0.120283 ⟩ = 1 ⟨ +1, +4, +7, +6 ≈ 20.311446 ⟩ × ⟨ 1, +4, 7, +6 ≈ 0.049233 ⟩ = 1 ⟨ +2, +2, 0, +1 ≈ 7.658328 ⟩ × ⟨ 4, +7, 2, 2 ≈ 0.130577 ⟩ = 1 ⟨ +2, +2, +1, 1 ≈ 7.597831 ⟩ × ⟨ +2, 2, 3, +5 ≈ 0.131617 ⟩ = 1 ⟨ +2, +3, +1, 1 ≈ 9.215865 ⟩ × ⟨ +1, 0, 1, 1 ≈ 0.108509 ⟩ = 1 ⟨ +2, +4, 0, 3 ≈ 8.422260 ⟩ × ⟨ 4, +5, +6, 12 ≈ 0.118733 ⟩ = 1 ⟨ +2, +4, +1, 2 ≈ 10.215865 ⟩ × ⟨ 1, +2, 0, 2 ≈ 0.097887 ⟩ = 1 ⟨ +2, +5, +2, 1 ≈ 13.627503 ⟩ × ⟨ +6, 7, 10, +19 ≈ 0.073381 ⟩ = 1 ⟨ +2, +5, +2, 3 ≈ 12.391435 ⟩ × ⟨ 2, +5, 2, 3 ≈ 0.080701 ⟩ = 1 ⟨ +2, +6, +1, 2 ≈ 13.451933 ⟩ × ⟨ 5, 6, +8, +16 ≈ 0.074339 ⟩ = 1 ⟨ +3, 0, +1, 1 ≈ 6.263876 ⟩ × ⟨ +18, 21, 29, +55 ≈ 0.159646 ⟩ = 1 ⟨ +3, 0, +1, +1 ≈ 7.499944 ⟩ × ⟨ +18, +21, 29, 55 ≈ 0.133334 ⟩ = 1 ⟨ +3, +3, 0, +1 ≈ 11.178475 ⟩ × ⟨ +3, +1, 3, 6 ≈ 0.089458 ⟩ = 1 ⟨ +3, +3, +2, 1 ≈ 12.293548 ⟩ × ⟨ 3, +3, +5, 8 ≈ 0.081343 ⟩ = 1 ⟨ +3, +5, 0, 4 ≈ 11.324373 ⟩ × ⟨ +3, 5, 0, +4 ≈ 0.088305 ⟩ = 1 ⟨ +3, +5, +1, 2 ≈ 13.736012 ⟩ × ⟨ +7, 13, +4, +5 ≈ 0.072801 ⟩ = 1 ⟨ +3, +5, +2, 2 ≈ 14.911582 ⟩ × ⟨ 1, 1, +2, +2 ≈ 0.067062 ⟩ = 1 ⟨ +3, +6, +2, 2 ≈ 16.529616 ⟩ × ⟨ 0, 0, 1, +2 ≈ 0.060497 ⟩ = 1 ⟨ +3, +6, +3, 1 ≈ 18.323221 ⟩ × ⟨ 0, 1, +3, 3 ≈ 0.054576 ⟩ = 1 ⟨ +3, +7, +3, 4 ≈ 18.087153 ⟩ × ⟨ 3, 5, +6, +11 ≈ 0.055288 ⟩ = 1 ⟨ +4, +1, +3, 1 ≈ 12.135164 ⟩ × ⟨ 29, 34, +47, +89 ≈ 0.082405 ⟩ = 1 ⟨ +4, +5, 0, +1 ≈ 16.316656 ⟩ × ⟨ 20, 23, +32, +61 ≈ 0.061287 ⟩ = 1 ⟨ +4, +7, 0, 6 ≈ 15.226486 ⟩ × ⟨ 20, +38, 12, 15 ≈ 0.065675 ⟩ = 1 ⟨ +4, +7, +2, 2 ≈ 20.049763 ⟩ × ⟨ 2, +2, 0, +1 ≈ 0.049876 ⟩ = 1 ⟨ +5, +5, +3, 2 ≈ 19.891379 ⟩ × ⟨ +5, 5, 8, +13 ≈ 0.050273 ⟩ = 1