Number theory, primitive pythagorean triples

[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.

 

[fresh_42]

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 $m\geq 3,n\geq 2$.

 

[Mathematicsresear]

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?

 

[fresh_42]

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 $m > n$.

 

[Mathematicsresear]

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.

 

[fresh_42]

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?

 

[tnich]

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.