Quintright reciprocals.
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Suppose:

• B is a quintright, and all four components are multiples of some integer n ≥ 2.
• C is any nonzero quintright.
• A is the product BC.

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
 Br Bq + Bs Br + Bt Bs 3Bq + Bs 2Br + Bt Bq + 2Bs Br + Bt Br + Bt Bq − Bt Bq − Bs Bs − 2Bq − Br − Bt Bq − 3Bs − Br − 2Bt

dq = det
 0 Bq + Bs Br + Bt Bs +1 2Br + Bt Bq + 2Bs Br + Bt 0 Bq − Bt Bq − Bs −1 − Br − Bt Bq − 3Bs − Br − 2Bt

dr = det
 Br 0 Br + Bt Bs 3Bq + Bs +1 Bq + 2Bs Br + Bt Br + Bt 0 − Bt Bq − Bs Bs − 2Bq −1 Bq − 3Bs − Br − 2Bt

ds = det
 Br Bq + Bs 0 Bs 3Bq + Bs 2Br + Bt +1 Br + Bt Br + Bt Bq 0 Bq − Bs Bs − 2Bq − Br − Bt −1 − Br − 2Bt

dt = det
 Br Bq + Bs Br + Bt 0 3Bq + Bs 2Br + Bt Bq + 2Bs +1 Br + Bt Bq − Bt 0 Bs − 2Bq − Br − Bt Bq − 3Bs −1

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

(BrBt)4

which for any integer (BrBt) 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
eveneveneveneven0
evenoddoddeven0
oddevenevenodd0
oddoddoddodd0
twelve other combinations1

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:

Agov (A)Bgov (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

• ⟨ −Aq, +Ar, −As, +At ⟩ × ⟨ −Bq, +Br, −Bs, +Bt ⟩ = 1
• ⟨ +Aq, −Ar, +As, −At ⟩ × ⟨ +Bq, −Br, +Bs, −Bt ⟩ = 1
• ⟨ +As, +At, −Aq, +Ar ⟩ × ⟨ +Bs, +Bt, −Bq, +Br ⟩ = 1

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
```