A correct answer will yield me an "A" all SEMESTER, so help is appreciated
Question:
Can you find a pythagorean triple whose nonhypotunese legs ARE NOT divisible by 12? That is to say that triples like 3,4,5 would not work because a(3)xb(4)=12/12=1 meaing that their product (a&b) is divisible by 12...
Your question is not clear
You must have meant to say:
..whose PRODUCT of the nonhypotunese legs is NOT divisible by 12..
That only becomes clear in your example when you first use the word product and show it in the example.
So far I have not found any.
I am working with this formula:
if m and n are whole numbers and m > n,
then 2mn, m^2 - n^2 and m^2 + n^2 will by Pythagorean triples.
Furthermore, if one of m or n is even and the other is odd, and they are relatively prime, then the triple will be in lowest terms.
What have you done so far?
Actually, I'm using the same formula as you. I searched for formulas yielding Pythagorean triples and it listed a few,yours being one. I'm not getting any results though..
To find a Pythagorean triple whose non-hypotenuse legs are not divisible by 12, we need to explore the concept of Pythagorean triples and their properties.
A Pythagorean triple consists of three positive integers (a, b, c), where a^2 + b^2 = c^2. In other words, the sum of the squares of the two shorter sides (legs) is equal to the square of the longest side (hypotenuse).
To approach this question, we can first find Pythagorean triples using a general formula. One such formula is:
a = m^2 - n^2
b = 2mn
c = m^2 + n^2
Here, m and n are positive integers, and m > n. By choosing different values for m and n, we can generate various Pythagorean triples.
Now, let's consider the divisibility condition. To ensure that the non-hypotenuse legs are not divisible by 12, we need to avoid combinations of (a, b) where either a or b is divisible by 12.
Starting with the formula, let's check for patterns and determine which values of m and n yield non-divisible legs.
For simplicity, let's assume m and n are coprime (having no common factors other than 1) so that a and b will also be coprime.
1. For a to be divisible by 12, both m^2 and n^2 must be divisible by 12.
- For m^2 to be divisible by 12, either m or m^2 must be divisible by 12.
- For n^2 to be divisible by 12, either n or n^2 must be divisible by 12.
2. However, since m and n are coprime, neither m nor n can be divisible by 12, as that would violate the coprime condition.
Based on this analysis, we can conclude that any values of m and n (both coprime) that satisfy the Pythagorean triple formula will result in non-divisible legs.
Therefore, any Pythagorean triple generated using the formula a = m^2 - n^2, b = 2mn, and c = m^2 + n^2, with m and n being coprime, will have non-hypotenuse legs that are not divisible by 12.
Example: Let's take m = 3 and n = 1.
Using the formula, we get:
a = 3^2 - 1^2 = 9 - 1 = 8
b = 2 * 3 * 1 = 6
c = 3^2 + 1^2 = 9 + 1 = 10
The resulting Pythagorean triple is (8, 6, 10), and neither 8 nor 6 is divisible by 12.
In conclusion, a Pythagorean triple whose non-hypotenuse legs are not divisible by 12 can be found using the formula a = m^2 - n^2, b = 2mn, and c = m^2 + n^2, where m and n are coprime.