The integer triangle problem

[Thelema]

The Triangle Problem

Suppose I construct a right angled triangle so that both sides and the hypotenuse are integers and all relatively prime to one another.

Prove that the area of ANY and all such triangles is not an odd product of the longest non-hypotenuse side.

edit: added all relatively prime to one another condition - forgot this. clarified that "longest side" refers to longest non-hypotenuse side.

Edited June 22, 2013 by Thelema

[ThunderIdeal]

okay if you explain to me how one does a mathematical proof, and any other math questions i have along the way.

[Wameron36]

A right angle triangle with perfect integer sides and hypotemus - Sides of length 3, 4 & 5 for example.

So the area would be: A = (3*4)/2 = 6

'... the area is not an odd product of the longest side.'

The area in this case is even, so no matter what it is a product of, it's still even... and is not an odd product... and not an odd product of the longest side (which is 5).

Other examples of these triangles:

sides of length 5, 12, 13 ----> A = (5*12)/2 = 30 .... An even number. So if it's a product of the longest side it's still an even number.

Sides of length 8, 15, 17 ----> A = (8*15)/2 = 60... which is even. And therefore an even product...

Not sure if that's what the whole idea is though???
If it is, not sure if just showing examples is proving it though.

[Wameron36]

An equation for finding the side lengths of triangles that have these properties:

m^2 - n^2 : 2mn : m^2 + n^2

Where: m and n are any perfect integer and m>n

For the side of length 2mn, any perfect integer chosen for m or n will result in 2mn being even for the fact that you're multiplying by 2, an even number, and hence the product, 2mn, has to be even.

Then to find the area, this side length is multiplied by the other side length (m^2 - n^2). Since 2mn is even, multiplying it by m^2 - n^2, we get an even product.

This is then divided by 2. Any even number divided by 2 will also be even. And therefore the area is even!

And since the area is even, whatever set of factors it is a product of, whether that's the longest side or anything else, it's still going to be an even product.

[CβL]

Suppose I construct a right angled triangle so that both sides and the hypotenuse are perfect integers (normal integer or perfect number?).

Prove that the area of the triangle is not an odd product (odd product means the hypotenuse multiplied by an odd integer?) of the longest side.

Edited June 14, 2013 by CβL

[CβL]

I try and frame these problems graphically. I've managed the equation
(AB)/(2*sqrt[A^2 + B^2]) = K

This is a 3D surface, with independent variables A and B, and dependent is K. I need to prove that at every grid-point (i.e. integers) that K is not an odd integer.

Wameron's answer is interesting (and uses a similar approach to many successful number theory proofs) - I'll think more about it.

[Thelema]

sorry, I should have mentioned, but I realised too late. All sides of the triangle have to be relatively prime to one another. And by "longest side" I mean the longest side that is NOT the hypotenuse. And examples won't do, I need a general proof.

I've adjusted the original post and repost here:

The Triangle Problem

Suppose I construct a right angled triangle so that both sides and the hypotenuse are integers, and that these integers are all relatively prime to one another.

Prove that the area of ANY and all such triangles is not an odd product of the longest side (not being the hypoteneuse).

Edited June 22, 2013 by Thelema

[Thelema]

An equation for finding the side lengths of triangles that have these properties:

m^2 - n^2 : 2mn : m^2 + n^2

Where: m and n are any perfect integer and m>n

For the side of length 2mn, any perfect integer chosen for m or n will result in 2mn being even for the fact that you're multiplying by 2, an even number, and hence the product, 2mn, has to be even.

Then to find the area, this side length is multiplied by the other side length (m^2 - n^2). Since 2mn is even, multiplying it by m^2 - n^2, we get an even product.

This is then divided by 2. Any even number divided by 2 will also be even. And therefore the area is even!

And since the area is even, whatever set of factors it is a product of, whether that's the longest side or anything else, it's still going to be an even product

.

Also, just because something is even and an even product of something does not mean it can't be the odd product of something else. eg 6 is even, and an even product, but it is still the odd product of 2. The question asks that the area cannot be represented as an odd number X 2mn.

Also, any even number divided by 2 will not also be even, it can be odd.. The formula is right,

so the question becomes:

prove that (2x +1)2mn = nm(m^2 - n^2) is false for all integer values of x.

Edited June 25, 2013 by Thelema

[Thelema]

OK this is easy now, the hard part was essentially deriving the formula that Wameron gave. Now we have prove that (2x +1)2mn = nm(m^2 - n^2) is false for all integer values of x.

Cancel nm from both

(2x + 1)2 = (m^2 -n^2)

Can you get it from there?