The Quaternion identity expresses the product of any two numbers, each expressed as the sum of four squares, as the sum of four squares.

(a2+b2+c2+d2)(A2+B2+C2+D2) = (aA+bB+cC+dD)2+(aB-bA-cD+dC)2+(aC+bD-cA-dB)2+(aD-bC+cB-dA)2

Derivation from quaternionsEdit

The quaternion units 1,i,j,k are defined so that

1 = 1*1 = -i2 = -j2 = -k2
i = 1i = i1 = jk = -kj
j = 1j = -ik = j1 = ki
k = 1k = ij = -ji = k1

If u=a-bi-cj-dk, U=A+Bi+Cj+Dk, then from


we get

(a2+b2+c2+d2)(A2+B2+C2+D2) = (aA+bB+cC+dD)2+(aB-bA-cD+dC)2+(aC+bD-cA-dB)2+(aD-bC+cB-dA)2

In defining u, the signs of b, c, and d are arbitrary; We could have just as easily defined them differently (or permuted the meanings of A, B, C and D). The important observations to make regarding the sign of the variables a,b,c,d,A,B,C,D on the right hand side of the equation are:

  • One of the eight variables, which we will arbitrarily call a has a positive sign whereever it appears. This fixes a as one of the lower-case variables.
  • One of the four variables upper-case variables, which we arbitrarily call A has a negative sign wherever it appears, except with a. This fixes the relationship between a and A.
  • Then the pairing of b, c, and d with B, C, and D is such that bC, cD, and dB have a negative sign; and cB, dC, and bD have a positive sign.


By defining $ u=a-bi-\sqrt{n}cj-\sqrt{n}dk\, $ and $ U=A+Bi+\sqrt{n}Cj+\sqrt{n}Dk\, $, then from


we get

(a2+b2+nc2+nd2)(A2+B2+nC2+nD2) =


A simple application of this identity is

3(A^2+B^2+C^2+D^2) = (A+B+C)^2+(D+A-B)^2+(C+D-A)^2+(B-C+D)^2

See alsoEdit

  • Complex product identity, that the product of two numbers expressed as the sum of two squares can be expressed as the sum of two squares.