using the primitive element of biquadratic field

Let m and n be two distinct squarefreeMathworldPlanetmath integers 1.  We want to express their square roots as polynomials of

α:=m+n (1)

with rational coefficients.

If α is cubed (, the result no terms with mn:

α3 =(m)3+3(m)2n+3m(n)2+(n)3

Thus, if we subtract from this the product (3m+n)α, the n term vanishes:


Dividing this equation by -2m+2n (0) yields

m=α3-(3m+n)α2(-m+n). (2)

Similarly, we have

n=α3-(m+3n)α2(m-n). (3)

The (2) and (3) may be interpreted as such polynomials as intended.

Multiplying the equations (2) and (3) we obtain a corresponding for the square root of mn which also lies in the quartic field  (m,n)=(m+n):


For example, in the special case  m:=2,n:=3  we have


Remark.  The sum (1) of two square roots of positive squarefree integers is always irrational, since in the contrary case, the equation (3) would say that n would be rational; this has been proven impossible here (

Title using the primitive element of biquadratic field
Canonical name UsingThePrimitiveElementOfBiquadraticField
Date of creation 2013-03-22 17:54:25
Last modified on 2013-03-22 17:54:25
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 12
Author pahio (2872)
Entry type Application
Classification msc 11R16
Synonym expressing two square roots with their sum
Synonym irrational sum of square roots
Related topic BinomialTheorem