Home page.

Many of the identities on the home page suggest defining the septright triad, which comes in two varieties: one for rtv3s and one for qsu3s. Let a, b, c, d, e be integers, and:

If X, Y, Z form a triad, so do these:

If most of the signs in the expression ± X ± Y ± Z are minus, it is canonical to negate all three of X, Y, Z.

Below are examples of triads, with addition or subtraction indicated.

QR· SV=+ 2R − 2T + 3V add
SV· TU=+ 2R − 3T + 2V add
TU· QR=+ 3R − 2T + 2V add
+ 7R − 7T + 7V result
7R÷ U=+ 3Q + 5S + 1U add
7T÷ S=+ 5Q − 1S − 3U add
7V÷ Q=+ 1Q − 3S + 5U sub
+ 7Q + 7S − 7U result
QT· RS=+ 4R − 1T + 2V add
RS· UV=+ 1R − 2T + 4V add
UV· QT=− 2R + 4T − 1V sub
+ 7R − 7T + 7V result
7R÷ Q=+ 1Q + 4S − 2U add
7T÷ U=+ 4Q + 2S − 1U add
7V÷ S=+ 2Q + 1S − 4U add
+ 7Q + 7S − 7U result
Q÷ U= 1 ÷ V=+ R + V add
U÷ S= 1 ÷ R=+ RT  add
S÷ Q= 1 ÷ T= + TV sub
+ 2R− 2T+ 2V result
Q÷ R=+ Q U add
S÷ V=+ Q+ S  add
U÷ T= + SU add
+ 2Q+ 2S− 2U result
T÷ R=− 1R+ 2T  add
V÷ T= + 1T− 2V add
R÷ V=+ 2R + 1V sub
− 3R+ 3T− 3V result
7T÷ Q= 7 ÷ S = − 1Q + 3S + 2U add
7V÷ U= 7 ÷ Q = + 3Q − 2S + 1U add
7R÷ S= 7 ÷ U = + 2Q + 1S + 3U sub
zero  result
R÷ T=+ 1R − 1T + 2V add
T÷ V=+ 2R − 1T + 1V add
V÷ R=− 1R + 2T − 1V sub
+ 4R − 4T + 4V result
S÷ T=+ QS + U add
U÷ R=Q + S + U add
Q÷ V=+ Q + S + U sub
QS + U result
A=+ 1R 5T + 5V add
B=+ 5R 1T + 5V add
C= 5R + 5T 1V sub
+ 11R11T + 11V result
D= 2R + 5T 4V add
F= 5R + 4T 2V add
E=+ 4R 2T + 5V sub
11R + 11T11V result

The triads in the bottom row come from section 8 of the home page.


Besides the sum of the elements of a triad, something might be said about their product.

Consider an rtv3 triad in the following form:

X′ = ⟨ 0, +a, 0, −b, 0, +c
Y′ = ⟨ 0, +b, 0, −c, 0, +a
Z′ = ⟨ 0, +c, 0, −a, 0, +b

Clearly, X′ + Y′ + Z′ = a + b + c. Further, X′Y′Z′ = d where d is a ten-term cubic polynomial:

d=abca3b3c3
3a2b − 3b2c − 3c2a
+ 4a2c + 4b2a + 4c2b

Now consider a qsu3 triad in the following form:

X″ = ⟨ +a, 0, −b, 0, +c, 0 ⟩
Y″ = ⟨ +b, 0, −c, 0, +a, 0 ⟩
Z″ = ⟨ +c, 0, −a, 0, +b, 0 ⟩

Of course, X″ + Y″ + Z″ = (a + b + c)√7. Also, X″Y″Z″ = e√7 where e is:

e=abca3b3c3
+ 2a2b + 2b2c + 2c2a
+ a2c + b2a + c2b