No, the multiple of any random triple is not a Pythagorean triple.
The set of integers represents the integers.
The set of positive odd integers.
The set of Counting Numbers or Natural Numbersincludes positive integers but not negative integers or zero.The set is 1,2,3,4,5,6....and so on.
Integers include negative numbers.
A Pythagorean triple is three positive integers a, b, and c, such that a^2 + b^2 = c^2. A well known Pythagorean triplet is (3,4,5). If (a, b, c) is a Pythagorean triplet, then so is (ka, kb, kc) for any positive integer k.
The integers 3 and 4 form part of the 3-4-5 Pythagorean triple.
There are infinitely many Pythagorean triples. To find a Pythagorean triple take two positive integers x, y with x > y. A Pythagorean triple is of the form x2 - y2, 2xy, x2 + y2.
If p and q are integers, then a = p2 - q2 b = 2pq, and c = p2 + q2 form a Pythagorean triple. Furthermore, if p and q are co-prime then the triple is primitive Pythagorean.
There are infinitely many combinations. Using any pair of integers m and n such that m > n > 0, the integers m2 - n2, 2mn and m2 + n2 will form a Pythagorean triple.
x=4 y=1
6, 8, and 10 is simply a scaled up version of a 3,4,5 triangle (simply double each side). Since 3,4,5 is a Pythagorean triple, so is the scaled up triangle. Alternatively, since 6, 8, and 10 are integers (whole numbers) that fulfill the Pythagorean theorem (62 + 82 = 102 ), they are a Pythagorean triple.
No, the multiple of any random triple is not a Pythagorean triple.
Nearly but not quite a Pythagorean triple
If you mean 3, 4 and 5 then yes it is a Pythagorean triple
A "Pythagorean triple" is a set of 3 numbers that fits the Pythagorean Theorem.So your question is: "Do 2, 6, and 8 form a Pythagorean triple ?", and the answeris either 'yes' or 'no'.(2)2 = 4(6)2 = 364 + 36 = 40(8)2 = 6440 is not equal to 64.So the answer is "no".
What is the value of x if (13, 84, x) is a Pythagorean triple?