On this blog I've discussed numbers of the form , which are intimately related to a mathematical construct known as the Gaussian integers, . Here we will explore another binary quadratic form.



Problem 1: (Engel, Chapter 6, #117) The product of two positive integers of the form has the same form.

Solution: We should recognize this expression from the sum-of-cubes expression. Consequently, let , a primitive third root of unity; then



This is analogous to the complex factorization . You may remember that I discussed the Gaussian integers in a previous post. Well, these are the Eisenstein integers, which also have unique prime factorization.

Anyway,

$\begin{eqnarray*}(a^{2}+ab+b^{2})(c^{2}+cd+d^{2}) &=& (a-b \omega)(a-b \omega^{2})(c-d \omega)(c-d \omega^{2}) \\ &=& (ac+bd \omega^{2}-(bc+ad) \omega)(ac+bd \omega-(bc+ad) \omega^{2}) \\ &=& ( (ac-bd)-(bc+ad+bd) \omega)( (ac-bd)-(bc+ad+bd) \omega^{2}) \\ &=& (ac-bd)^{2}+(ac-bd)(bc+ad+bd)+(bc+ad+bd)^{2}\\ \end{eqnarray*}$

(I have used the identity .) Knowing the analogue for the Gaussian integers makes working through the algebra that much easier. QED.

Remark: Every Eisenstein integer can be expressed in the form owing to the identity I used. The multiplicative norm (as we have shown here) is ; indeed, is the complex conjugate of . We can prove the above result using an argument based on the fact that conjugation is multiplicative here too.

Problem 2: (Engel, Chapter 6, #127) Find infinitely many integer solutions to



Solution: Engel gives a simple solution, so I will give a complicated one. We know from Problem 1 that the LHS is equal to



Clearly one easy subset of solutions occurs when . Indeed, this gives the identity



Interestingly enough, there are two choices for . The second choice is a somewhat uncommon factorization; it can be derived by difference of squares.

Problem 3: (Engel, Chapter 6, #136) Show that if is prime, where , then is a power of .

Solution: The factorization derived in Problem 2 suggests the possibility that



Is it true? Well, a polynomial divides another iff the roots of the divisor are all roots of the dividend; those roots are the roots of unity . Plugging in, we desire that



In other words, if has any prime factors which are not then



Hence has no such factors. QED.

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Practice Problem: (Engel, Chapter 6, #118) Given , show that if has one rational solution then it has infinitely many.