You can generate a Pythagorean triple using the following "formula"
take any values of m and n, where m > n
then
2mn
m^2 - n^2
m^2 + n^2
will form a Pythagorean triple
e.g. m = 6 , n = 5
2mn = 60
m^2-n^2 = 11
m^2 + n^2 = 61
and 60^2 + 11^2 = 61^2
so 11, 60, 61 will form a Pythagorean triple
if you pick both m and n as even, your triple will be even
btw, if you choose m and n so that they are relatively prime, then you will get a triple in lowest terms.
e.g. 6,8,10 is 2 times the 3,4,5, which is the smallest triple that exists.
So by multiplying any triple by an even factor will yield a triple of even numbers.
As to your question of 3 consecutive even numbers
being Pytagorean triples, the 6,8,10 is the only one
I wanted to know a pythagorean example for three consecutive even numbers like 6,8,10 . I want a new example,
2 answers
I have not checked or proved, but believe (6,8,10) is the only triplet that is consecutive even.
If you would like to search for more triplets, here's a reference to the work of Brahmagupta:
"Pythagorean triples[edit]
In chapter twelve of his Brahmasphutasiddhanta, Brahmagupta provides a formula useful for generating Pythagorean triples:
12.39. The height of a mountain multiplied by a given multiplier is the distance to a city; it is not erased. When it is divided by the multiplier increased by two it is the leap of one of the two who make the same journey.[14]
Or, in other words, if d = mx/(x + 2), then a traveller who "leaps" vertically upwards a distance d from the top of a mountain of height m, and then travels in a straight line to a city at a horizontal distance mx from the base of the mountain, travels the same distance as one who descends vertically down the mountain and then travels along the horizontal to the city.[14] Stated geometrically, this says that if a right-angled triangle has a base of length a = mx and altitude of length b = m + d, then the length, c, of its hypotenuse is given by c = m (1+x) – d. And, indeed, elementary algebraic manipulation shows that a2 + b2 = c2 whenever d has the value stated. Also, if m and x are rational, so are d, a, b and c. A Pythagorean triple can therefore be obtained from a, b and c by multiplying each of them by the least common multiple of their denominators."
from:
http://en.wikipedia.org/wiki/Brahmagupta
If you would like to search for more triplets, here's a reference to the work of Brahmagupta:
"Pythagorean triples[edit]
In chapter twelve of his Brahmasphutasiddhanta, Brahmagupta provides a formula useful for generating Pythagorean triples:
12.39. The height of a mountain multiplied by a given multiplier is the distance to a city; it is not erased. When it is divided by the multiplier increased by two it is the leap of one of the two who make the same journey.[14]
Or, in other words, if d = mx/(x + 2), then a traveller who "leaps" vertically upwards a distance d from the top of a mountain of height m, and then travels in a straight line to a city at a horizontal distance mx from the base of the mountain, travels the same distance as one who descends vertically down the mountain and then travels along the horizontal to the city.[14] Stated geometrically, this says that if a right-angled triangle has a base of length a = mx and altitude of length b = m + d, then the length, c, of its hypotenuse is given by c = m (1+x) – d. And, indeed, elementary algebraic manipulation shows that a2 + b2 = c2 whenever d has the value stated. Also, if m and x are rational, so are d, a, b and c. A Pythagorean triple can therefore be obtained from a, b and c by multiplying each of them by the least common multiple of their denominators."
from:
http://en.wikipedia.org/wiki/Brahmagupta
2 answers