The Function x^2 + xy + y^2 – Part 9

9)      The product of two numbers in the form of x² + xy + y² is another number in the form of x² + xy + y², making x² + xy + y² closed in multiplication.

The product of two numbers in the form of x² + xy + y² is another number in the form of x² + xy + y².  For example, (1, 2) ∙ (1, 4) = (7, 7) because (1, 2) = 1² + 1∙2 + 2² = 7 and (1, 4) = 1² + 1∙4 + 4² = 21 and (7, 7) = 7² + 7∙7 + 7² = 147, and 7 ∙ 21 = 147.

	1	2	3	4	5	6	7	8	9	10
1	3	7	13	21	31	43	57	73	91	111
2	7	12	19	28	39	52	67	84	103	124
3	13	19	27	37	49	63	79	97	117	139
4	21	28	37	48	61	76	93	112	133	156
5	31	39	49	61	75	91	109	129	151	175
6	43	52	63	76	91	108	127	148	171	196
7	57	67	79	93	109	127	147	169	193	219
8	73	84	97	112	129	148	169	192	217	244
9	91	103	117	133	151	171	193	217	243	271
10	111	124	139	156	175	196	219	244	271	300

We can arrive at a general format for this by looking at (a, b) ∙ (c, d) = (e, f) and finding values for e and f in terms of a, b, c, and d.  We know that (a, b) ∙ (c, d) = (a² + ab + b²)(c² + cd + d²) = a²c² + a²cd + a²d² + abc² + abcd + abd² + b²c² + b²cd + b²d² = e² + ef + f².  If e and f are in terms of a, b, c, and d, it is likely that each term has a degree of 2 so that e², ef, and f² combine to make all terms with a degree of 4.  Also, it is likely that there is no a² term in e and f because there is no a⁴ term in a²c² + a²cd + a²d² + abc² + abcd + abd² + b²c² + b²cd + b²d².  (If there were, say e = (ma² + …) and f = (na² + …) then e² + ef + f² = (ma² + …)² + (ma² + …)(na² + …) + (na² + …)² = m²a⁴ + mna⁴ + n²a⁴ + …. = (m² + mn + n²)a⁴ + … = (0a⁴ + …), which means m² + mn + n² = 0.  But if m and n are integers, which is also likely, then m = 0 and n = 0, which means there is no a² term in e or f.)  The same argument can be made for no b², c², d², ab, or cd term.  (It’s not necessarily true, but if it works it will cut the work down in half.)  So we can let e = m₁ac + m₂ad + m₃bc + m₄bd and f = n₁ac + n₂ad + n₃bc + n₄bd, which means e² + ef + f² = (m₁ac + m₂ad + m₃bc + m₄bd)² + (m₁ac + m₂ad + m₃bc + m₄bd)(n₁ac + n₂ad + n₃bc + n₄bd) + (n₁ac + n₂ad + n₃bc + n₄bd)².  Using charts to help multiply:

	a2	ab	b2
c2	a2c2	abc2	b2d2
cd	a2cd	abcd	b2cd	=
d2	a2d2	abd2	b2d2

		m1ac	m2ad	m3bc	m4bd
	m1ac	m12a2c2	m1m2a2cd	m1m3abc2	m1m4abcd
	m2ad	m1m2a2cd	m22a2d2	m2m3abcd	m2m4abd2	+
	m3bc	m1m3abc2	m2m3abcd	m32b2c2	m3m4b2cd
	m4bd	m1m4abcd	m2m4abd2	m3m4b2cd	m42b2d2

		m1ac	m2ad	m3bc	m4bd
	n1ac	m1n1a2c2	m2n1a2cd	m3n1abc2	m4n1abcd
	n2ad	m1n2a2cd	m2n2a2d2	m3n2abcd	m4n2abd2	+
	n3bc	m1n3abc2	m2n3abcd	m3n3b2c2	m4n3b2cd
	n4bd	m1n4abcd	m2n4abd2	m3n4b2cd	m4n4b2d2

		n1ac	n2ad	n3bc	n4bd
	n1ac	n12a2c2	n1n2a2cd	n1n3abc2	n1n4abcd
	n2ad	n1n2a2cd	n22a2d2	n2n3abcd	n2n4abd2
	n3bc	n1n3abc2	n2n3abcd	n32b2c2	n3n4b2cd
	n4bd	n1n4abcd	n2n4abd2	n3n4b2cd	n42b2d2

This gives us 9 new formulas:

for the a²c² term: 1 = m₁² + m₁n₁ + n₁²

for the a²d² term: 1 = m₂² + m₂n₂ + n₂²

for the b²c² term: 1 = m₃² + m₃n₃ + n₃²

for the b²d² term: 1 = m₄² + m₄n₄ + n₄²

for the abc² term: 1 = 2m₁m₃ + m₃n₁ + m₁n₃ + 2n₁n₃

for the a²cd term: 1 = 2m₁m₂ + m₂n₁ + m₁n₂ + 2n₁n₂

for the b²cd term: 1 = 2m₃m₄ + m₃n₄ + m₄n₃ + 2n₃n₄

for the abd² term: 1 = 2m₂m₄ + m₂n₄ + m₄n₂ + 2n₂n₄

for the abcd term: 1 = 2m₁m₄ + 2m₂m₃ + m₁n₄ + m₂n₃ + m₃n₂ + m₄n₁ + 2n₁n₄ + 2n₂n₃

There are 6 possible integer answers for the a²c² formula 1 = m₁² + m₁n₁ + n₁² which are (m₁, n₁) = (0, 1), (1, 0), (0, -1), (-1, 0), (1, -1), or (-1, 1).  Since these are all related, we can choose one for now and derive the others based on the equality formulas from Point #7.  So let’s choose (m₁, n₁) = (0, 1). 

Using the abc² term formula 1 = 2m₁m₃ + m₃n₁ + m₁n₃ + 2n₁n₃, 1 = 2(0)m₃ + m₃(1) + (0)n₃ + 2(1)n₃ or 1 = m₃ + 2n₃.  But since in the b²c² term formula 1 = m₃² + m₃n₃ + n₃² which means (m₃, n₃) = (0, 1), (1, 0), (0, -1), (-1, 0), (1, -1), or (-1, 1), the only two that work in 1 = m₃ + 2n₃ are (m₃, n₃) = (1, 0) or (-1, 1).  Again, these are all related, so we can choose (m₃, n₃) = (1, 0) and derive the other based on the equality formulas from Point #7.  So if (m₃, n₃) = (1, 0), then using the b²cd term formula 1 = 2m₃m₄ + m₃n₄ + m₄n₃ + 2n₃n₄, 1 = 2(1)m₄ + (1)n₄ + m₄(0) + 2(0)n₄ or 1 = 2m₄ + n₄.  But since in the b²d² term formula 1 = m₄² + m₄n₄ + n₄² which means (m₄, n₄) = (0, 1), (1, 0), (0, -1), (-1, 0), (1, -1), or (-1, 1), the only two that work in 1 = 2m₄ + n₄ are (m₄, n₄) = (0, 1) or (1, -1). 

Similarly, if (m₁, n₁) = (0, 1), using the a²cd term formula and the a²d² term formula, (m₂, n₂) = (1, 0) or (-1, 1) and using the abd² term formula if (m₂, n₂) = (1, 0) then (m₄, n₄) = (0, 1) or (1, -1). 

Finally using the abcd term formula, 1 = 2m₁m₄ + 2m₂m₃ + m₁n₄ + m₂n₃ + m₃n₂ + m₄n₁ + 2n₁n₄ + 2n₂n₃, 1 = 2(0)m₄ + 2(1)(1) + (0)n₄ + (1)(0) + (1)(0) + m₄(1) + 2(1)n₄ + 2(0)(0) or 1 = 2 + m₄ + 2n₄.  Then (m₄, n₄) = (0, 1) is not a solution but (m₄, n₄) = (1, -1) is.

This means that one solution for (m₁, m₂, m₃, m₄, n₁, n₂, n₃, n₄) is (0, 1, 1, 1, 1, 0, 0, -1), which means that one solution for (e, f) is (ad + bc + bd, ac – bd).  We can test this with our original example of (1, 2) ∙ (1, 4) = (7, 7), because if a = 1, b = 2, c = 1, and d = 4, then e = ad + bc + bd = 1∙4 + 2∙1 + 2∙4 = 14 and f = ac – bd = 1∙1 – 2∙4 = -7, and (14, -7) = (7, 7) using the equivalency statement (x, y) = (-y, x + y) in Point #7.  In fact, we can use the equivalence statements on a and b, c and d, and e and f and arrive at 24 possible solutions for e and f:

1)      (a, b) ∙ (c, d) = (e, f) = (ad + bc + bd, ac – bd)

2)      (a, b) ∙ (c, d) = (e, f) = (ad + bc + bd, -ac – ad – bc)

3)      (a, b) ∙ (c, d) = (e, f) = (ac – bd, -ac – ad – bc)

4)      (a, b) ∙ (c, d) = (e, f) = (ac – bd, ad + bc + bd)

5)      (a, b) ∙ (c, d) = (e, f) = (-ac – ad – bc, ad + bc + bd)

6)      (a, b) ∙ (c, d) = (e, f) = (-ac – ad – bc, ac – bd)

7)      (a, b) ∙ (c, d) = (e, f) = (-ad – bc – bd, -ac + bd)

8)      (a, b) ∙ (c, d) = (e, f) = (-ad – bc – bd, ac + ad + bc)

9)      (a, b) ∙ (c, d) = (e, f) = (-ac + bd, ac + ad + bc)

10)   (a, b) ∙ (c, d) = (e, f) = (-ac + bd, -ad – bc – bd)

11)   (a, b) ∙ (c, d) = (e, f) = (ac + ad + bc, -ad – bc – bd)

12)   (a, b) ∙ (c, d) = (e, f) = (ac + ad + bc, -ac + bd)

13)   (a, b) ∙ (c, d) = (e, f) = (ac + bd + bc, ad – bc)

14)   (a, b) ∙ (c, d) = (e, f) = (ac + bd + bc, -ad – ac – bd)

15)   (a, b) ∙ (c, d) = (e, f) = (ad – bc, -ad – ac – bd)

16)   (a, b) ∙ (c, d) = (e, f) = (ad – bc, ac + bd + bc)

17)   (a, b) ∙ (c, d) = (e, f) = (-ad – ac – bd, ac + bd + bc)

18)   (a, b) ∙ (c, d) = (e, f) = (-ad – ac – bd, ad – bc)

19)   (a, b) ∙ (c, d) = (e, f) = (-ac – bd – bc, -ad + bc)

20)   (a, b) ∙ (c, d) = (e, f) = (-ac – bd – bc, ad + ac + bd)

21)   (a, b) ∙ (c, d) = (e, f) = (-ad + bc, ad + ac + bd)

22)   (a, b) ∙ (c, d) = (e, f) = (-ad + bc, -ac – bd – bc)

23)   (a, b) ∙ (c, d) = (e, f) = (ad + ac + bd, -ac – bd – bc)

24)   (a, b) ∙ (c, d) = (e, f) = (ad + ac + bd, -ad + bc)

In fact, we can use the same argument to prove that the product of two numbers in the form of x² + kxy + y² is another number in the form of x² + kxy + y² for any integer of k, making x² + kxy + y² closed in multiplication.  It turns out that (a, b) ∙ (c, d) = (e, f) = (ad + bc + kbd, ac – bd), along with 23 other solutions. The function x² + xy + y² is just a special case for when k = 1.