# Number theory, primitive pythagorean triples

• Mathematicsresear
In summary: There are only 8 ways to factor 120 into the product of two positive integers. Maybe you could just try them.
Mathematicsresear

## Homework Statement

Find the known pythagorean triangles with sides of integers lengths, given the area of the pythagorean triangle is 60.

## Homework Equations

Pythagorean triangles are right angled triangles.
a primitive pythagorean triple is of the form:

x=2mn
y=m^2-n^2
z=m^2+n^2

gcd(m,n)=1 so one is even and the other is odd.[/B]

## The Attempt at a Solution

xy=120 so d^2ab =120
( d I x = a , d I y = b)[/B]
the only possible values for d^2 are 1 and 4
so I have two cases:

1) ab = 120
2) ab=30

I can now write a = 2mn
b= m^2-n^2

so mn(m^2-n^2)=15
mn(m-n)(m+n)=15

I am unsure as what to do next.

Mathematicsresear said:

## Homework Statement

Find the known pythagorean triangles with sides of integers lengths, given the area of the pythagorean triangle is 60.

## Homework Equations

Pythagorean triangles are right angled triangles.
a primitive pythagorean triple is of the form:

x=2mn
y=m^2-n^2
z=m^2+n^2

gcd(m,n)=1 so one is even and the other is odd.[/B]

## The Attempt at a Solution

xy=120 so d^2ab =120
( d I x = a , d I y = b)[/B]
the only possible values for d^2 are 1 and 4
so I have two cases:

1) ab = 120
2) ab=30

I can now write a = 2mn
b= m^2-n^2

so mn(m^2-n^2)=15
mn(m-n)(m+n)=15

I am unsure as what to do next.
What do you need ##d## for, if you end up at ##mn(m^2-n^2)=c## anyway? We know from the start that ##mn(m^2-n^2)=60##. So ##n=1## gives an easy solution, and we're left with ##m\geq 3,n\geq 2##.

fresh_42 said:
What do you need ##d## for, if you end up at ##mn(m^2-n^2)=c## anyway? We know from the start that ##mn(m^2-n^2)=60##. So ##n=1## gives an easy solution, and we're left with ##m\geq 3,n\geq 2##.
How did you get m is greater than or equal to 3 and n is greater than or equal to 2?

Mathematicsresear said:
How did you get m is greater than or equal to 3 and n is greater than or equal to 2?
I first considered the case ##n=1##. For the rest we thus have ##n\geq 2##. Now I assume we have a real triangle, so ##y>0## which means ##m > n##.

fresh_42 said:
I first considered the case ##n=1##. For the rest we thus have ##n\geq 2##. Now I assume we have a real triangle, so ##y>0## which means ##m > n##.
Alright, I am told that this equality holds, but I am not sure how, mn(m-n)(m+n) >= mn(m+t) >= 24 but I am not sure how to interpret this answer, and where the 24 came from.

Mathematicsresear said:
Alright, I am told that this equality holds, but I am not sure how, mn(m-n)(m+n) >= mn(m+t) >= 24 but I am not sure how to interpret this answer, and where the 24 came from.
What 24? I could follow you until your cases 1) and 2) and then I got lost. But I think you don't need neither ##(m,n)=1## nor ##d##. The text says ##\frac{1}{2}xy=60=mn(m+n)(m-n)##. For ##n=1## we have the product of three consecutive numbers: ##60=(m-1)\cdot m \cdot (m+1)## which leaves not much room for solutions, because ##m \in \mathbb{N}##. Now we have the other case ##m>n\geq 2##. Does this lead to other solutions?

Mathematicsresear said:

## Homework Statement

Find the known pythagorean triangles with sides of integers lengths, given the area of the pythagorean triangle is 60.

## Homework Equations

Pythagorean triangles are right angled triangles.
a primitive pythagorean triple is of the form:

x=2mn
y=m^2-n^2
z=m^2+n^2

gcd(m,n)=1 so one is even and the other is odd.[/B]

## The Attempt at a Solution

xy=120 so d^2ab =120
( d I x = a , d I y = b)[/B]
the only possible values for d^2 are 1 and 4
so I have two cases:

1) ab = 120
2) ab=30

I can now write a = 2mn
b= m^2-n^2

so mn(m^2-n^2)=15
mn(m-n)(m+n)=15

I am unsure as what to do next.
There are only 8 ways to factor 120 into the product of two positive integers. Maybe you could just try them.

member 587159

## 1. What is number theory?

Number theory is a branch of mathematics that deals with the properties and relationships of numbers. It involves studying integers, prime numbers, and their patterns and structures.

## 2. What are primitive Pythagorean triples?

Primitive Pythagorean triples are sets of three positive integers (a, b, c) that satisfy the Pythagorean theorem (a^2 + b^2 = c^2) and have no common factors. In other words, they are the smallest possible sets of positive integers that form a right triangle.

## 3. How do you find primitive Pythagorean triples?

There are various methods for finding primitive Pythagorean triples, such as using the Euclid's formula (a = m^2 - n^2, b = 2mn, c = m^2 + n^2) or the Fibonacci sequence. These methods involve manipulating numbers to satisfy the Pythagorean theorem and ensure that the resulting triple is primitive.

## 4. What is the significance of primitive Pythagorean triples?

Primitive Pythagorean triples have been studied for centuries and have many applications in mathematics and beyond. They have connections to number theory, geometry, and even cryptography. They also provide insight into the fundamental properties of integers and their relationships.

## 5. Are there an infinite number of primitive Pythagorean triples?

Yes, there are an infinite number of primitive Pythagorean triples. This was proven by the Greek mathematician Euclid in his work "Elements" over 2000 years ago. However, finding them can be challenging, as there is no known formula or pattern for generating all possible primitive Pythagorean triples.

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