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:
〈 Bq, Br, Bs, Bt 〉 × 〈 Cq, Cr, Cs, Ct 〉 = 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 | = Cq(Br) | + Cr(Bq + Bs) | + Cs(Br + Bt) | + Ct(Bs) |
+1 | = Cq(3Bq + Bs) | + Cr(2Br + Bt) | + Cs(Bq + 2Bs) | + Ct(Br + Bt) |
0 | = Cq(Br + Bt) | + Cr(Bq) | + Cs(− Bt) | + Ct(Bq − Bs) |
−1 | = Cq(Bs − 2Bq) | + Cr(− Br − Bt) | + Cs(Bq − 3Bs) | + Ct(− Br − 2Bt) |
As before, regard this as a system of four linear equations in the four unknowns Cq, Cr, Cs, Ct 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:
du = det |
|
dq = det |
|
dr = det |
|
ds = det |
|
dt = det |
|
Then the quintright C = 1 ÷ B = 〈 dq ÷ du, dr ÷ du, ds ÷ du, dt ÷ du 〉 exists whenever du ≠ 0 and each of the four component division operations yields no remainder. If du 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 du. Expressed as a polynomial, du expands to an eighteen-term quartic form, in other words a polynomial where every term is of exactly the same degree:
+ 5Bq4 + 1Br4 + 5Bs4 + 1Bt4 |
− 10Bq3Bs + 6Br3Bt + 10BqBs3 + 6BrBt3 |
− 5Bq2Br2 − 5Bq2Bs2 − 10Bq2Bt2 − 10Br2Bs2 + 11Br2Bt2 − 5Bs2Bt2 |
− 10Bq2BrBt + 10BqBr2Bs − 10BrBs2Bt − 10BqBsBt2 |
Now consider du in modulo 5, as obtained by discarding obvious multiples of 5:
1Br4 + 6Br3Bt + 11Br2Bt2 + 6BrBt3 + 1Bt4
Subtract 10Br3Bt + 5Br2Bt2 + 10BrBt3 (which is a multiple of five) in order to yield:
1Br4 − 4Br3Bt + 6Br2Bt2 − 4BrBt3 + 1Bt4
which factors into
(Br − Bt)4
which for any integer (Br − Bt) equals 0 or 1 in modulo 5. Hence du cannot equal −1.
Something can also be said about the behavior of du in modulo 4. Each of Bq, Br, Bs, and Bt is even or odd, making 16 = 24 combinations, and each of these is easy to investigate separately. It turns out that in modulo 4, du always equals 0 or 1, as detailed in the following table:
Bq | Br | Bs | Bt | du 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 du must equal either 0 or 1 in each of modulos 4 and 5, then du must equal 0, 1, 5, or 16 in modulo 20.
Noteworthy is that du is preserved under multiplication. Consider any two quintrights A and B. Let the symbol du(B) represent du as calculated on Bq, Br, Bs, and Bt, exactly as above. Symbol du(A) works the same way. Then du(AB) = du(A) du(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 almost-reciprocal 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 (dq, dr, ds, dt) × lcm (dq, dr, ds, dt, du) ÷ lcm (dq, dr, ds, dt)
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 〈 Aq, Ar, As, At 〉 × 〈 Bq, Br, Bs, Bt 〉 = 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