Pythagorean Multiples from the Diophantine Equation x² ± D(∑ y_j²) = z² (Part XII)

A Method of Generating Multiples from Novel Equations

Previous sections, e.g., Part III and Part III produced Diophantine triples and quadruples using the Diophantine equation x² + D(∑ y_j²) = z². This section will show that completely integral and partially integral values of X, Y and Z are possible by this method.

The first Pythagorean partially integral triplet uses Table I where all three a_j = 1:

Table I (D = 1, n= 1)
			x	y₁	y₂	y₃	z
a₁	a₂	a₃	D(a₁² + a₂² + a₃²) − n²	2na₁	2na₂	2na₃	D(a₁² + a₂² + a₃²) + n²
1	1	1	(1 + 1 + 1) − 1 = 2	2	2	2	(1 + 1 + 1) + 1 = 4
squares			4	4	4	4	16
sum y_j²s			4	12			16
factor out			1	3			4
square roots			1	√3			2

Consequently, this method produces the 30^°,60^°,90^° triangle. Inspection of the "factor out" row, in addition, shows that the legs l₁ is odd l₂ is the square root of an odd number while the hypotenuse is even, contrary to (l₁ being odd when l₂ is even or l₁ being even when l₂ is odd) and the hypotenuse being odd. In addition, after factoring out 4, both x² and y² are odd.

Table II shows the result when using one a_j as the generating number:

Table II (D = 1, n= 1)
	x	y	z
a₁	D(a₁²) − n²	2na₁	D(a₁)² + n²
3	(3²) − 1 = 8	6	(3²) + 1 = 10
squares	64	36	100
factor out	16	9	25
square roots	4	3	5

giving us the 3,4,5 triangle. Table III employs three a_j numbers similar to Table I but different in the odd/even square numbers that are generated:

Table III (D = 1, n= 1)
			x	y₁	y₂	y₃	z
a₁	a₂	a₃	D(a₁² + a₂² + a₃²) − n²	2na₁	2na₂	2na₃	D(a₁² + a₂² + a₃²) + n²
2	2	2	(4 + 4 + 4) − 1 = 11	4	4	4	(4 + 4 + 4) + 1 = 13
squares			121	16	16	16	169
sum y_j²s			121	48			169
factor out			-	-			-
square roots			11	√48			13

where the x² is odd and the sum of y² is even. Table IV uses a D of 2 and one a_j = 5 and again where the total y² is equal to 2(y₁² + y₂² + y₃²):

Table IV (D = 2, n= 1)
			x	y₁	y₂	y₃	z
a₁	a₂	a₃	D(a₁² + a₂² + a₃²) − n²	2na₁	2na₂	2na₃	D(a₁² + a₂² + a₃²) + n²
5	1	1	2(25 + 1 + 1) − 1 = 53	10	2	2	2(25 + 1 + 1) + 1 = 55
squares			2809	100	4	4	3025
sum y_j²s			2809	216			3025
factor out			-	-			-
square roots			53	√216			55

and where the x² is odd and the sum of y² is even.

This concludes Part XII. The next section (Part XIII) will show how to generate the 1,1,√2 triplet using a modified form of the current method since the current method doesn't allow a way to produce it.
While (Part XIV) will show a novel way of obtaining sequences of fully integral Pythagorean triples.

Go back to Part XI. Go back to homepage.

Pythagorean Multiples from the Diophantine Equation x2 ± D(∑ yj2) = z2 (Part XII)

A Method of Generating Multiples from Novel Equations

Pythagorean Multiples from the Diophantine Equation x² ± D(∑ y_j²) = z² (Part XII)