I wanted to know a pythagorean example for three consecutive even numbers like 6,8,10 . I want a new example,
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 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
To find a new example of three consecutive even numbers that satisfy the Pythagorean theorem, let's follow a step-by-step process:
Step 1: Choose the first even number.
Let's start with any even number of your choice. For this example, let's take 12 as the first even number.
Step 2: Determine the subsequent even numbers.
To find the next two consecutive even numbers, add 2 to the first even number chosen in Step 1. In this case, 12 + 2 = 14. Then, add 2 again to get the third even number: 14 + 2 = 16.
Step 3: Apply the Pythagorean theorem.
The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, we can create a right-angled triangle using the three consecutive even numbers: 12, 14, and 16. Let's assign these numbers to the sides of the triangle.
The smallest side will be the one that is opposite the right angle (let's call it a), the slightly longer side will be adjacent to the right angle (let's call it b), and the longest side will be the hypotenuse (let's call it c).
So, in our triangle, we have:
Side a = 12
Side b = 14
Side c (hypotenuse) = 16
Now, let's check if these three sides satisfy the Pythagorean theorem:
a² + b² = c²
12² + 14² = 16²
144 + 196 = 256
340 = 256
Since the equation is not true (340 is not equal to 256), these particular numbers (12, 14, 16) do not form a Pythagorean triple.
To find another example, repeat steps 1-3 with different starting even numbers until you find a set of numbers that satisfies the Pythagorean theorem (where the sum of the squares of the smaller sides equals the square of the longest side).