1. ## Question 14

We have two problems this week.
First one comes from CaptainBlank.

1)Show that if $N$ has $p$ divisors (including 1 and itself),
that the product of all of these divisors is $\sqrt{N^p}$.

2)A "Pythagorean Triangle" is a triangle with integer sides. For example, 3-4-5. Show that the area of a Pythagorean Triangle is never a square. (Fermat)

2. Originally Posted by ThePerfectHacker
We have two problems this week.
First one comes from CaptainBlank.

...

2)A "Pythagorean Triangle" is a triangle with integer sides. For example, 3-4-5. Show that the area of a Pythagorean Triangle is never a square. (Fermat)
Hello,

to 2): (I don' know how to "whiten" the Latex formulae, therefore I post my reply completely visible)

A right triangle with hypotenuse c and legs a and b has integer sides if
a = u&#178; - v&#178;, u > v and u, v in IN
b = 2uv
c = u&#178; + v&#178;

(for instance u = 2, v = 1 will give the famous 3, 4, 5 triangle)

The area of a right triangle is
$A=\frac{a \cdot b}{2}$

Plug in the above mentioned terms for a and b:

$A=\frac{(u^2+v^2) \cdot 2uv}{2}=(u^2-v^2) \cdot uv$

Now you have to prove that
- $u^2-v^2=(u+v)(u-v)$ is only a square if u=v or u=1 and v=0
- $u \cdot v$ is only a square if u = v
- $u^2-v^2=uv$. Solve this equation for u and you'll get:
$u=\frac{v \pm v \cdot \sqrt{5}}{2}$ that means u is not in IN

Thus the area can't be a square.

EB

Second thought:

$A=(u^2-v^2)uv=(u+v)(u-v)\cdot u \cdot v$
If this is a square there must be two pairs of equal factors which isn't possible under the given conditions. Thus A can't be a square.

3. Originally Posted by earboth
Hello,

to 2): (I don' know how to "whiten" the Latex formulae, therefore I post my reply completely visible)

A right triangle with hypotenuse c and legs a and b has integer sides if
a = uČ - vČ, u > v and u, v in IN
b = 2uv
c = uČ + vČ

(for instance u = 2, v = 1 will give the famous 3, 4, 5 triangle)

The area of a right triangle is
$A=\frac{a \cdot b}{2}$

Plug in the above mentioned terms for a and b:

$A=\frac{(u^2+v^2) \cdot 2uv}{2}=(u^2-v^2) \cdot uv$

Now you have to prove that
- $u^2-v^2=(u+v)(u-v)$ is only a square if u=v or u=1 and v=0
- $u \cdot v$ is only a square if u = v
- $u^2-v^2=uv$. Solve this equation for u and you'll get:
$u=\frac{v \pm v \cdot \sqrt{5}}{2}$ that means u is not in IN

Thus the area can't be a square.

EB
As defined by TPH, a "Pythagorean triangle" need only have integer length sides, it doesn't have to include a right angle.

-Dan

4. Here's how you do it, earboth. Click on quote to see

highlight between *earboth*

5. Originally Posted by topsquark
As defined by TPH, a "Pythagorean triangle" need only have integer length sides, it doesn't have to include a right angle.
That was a mistake by me. I assumed when I said "Pythagorean" it will be taken as right.

6. Originally Posted by galactus
Here's how you do it, earboth. Click on quote to see

highlight between *earboth*
Hello, galactus,

thanks a lot, but the problem I couldn't master is:

Hi, here you have an equation written in Latex: $y = x^2$ and this isn't whitened

EB

7. Originally Posted by earboth
Hello, galactus,

thanks a lot, but the problem I couldn't master is:

Hi, here you have an equation written in Latex: $y = x^2$ and this isn't whitened

EB
It's a know problem (at least to some of us) that I have struggled with
in the past, and not found a solution short of not using LaTeX in invisible
text .

RonL

8. Originally Posted by ThePerfectHacker
That was a mistake by me. I assumed when I said "Pythagorean" it will be taken as right.
Well that makes the problem a lot easier and explains why I couldn't format a proof!

-Dan

9. I was looking at earboth's approach but I did not understand it. For example if $uv$ is a square only if $u=v$ but that is not true, what happens when $u=4,v=9$?
-----
The "official" answer appeared in the note of a book owned by Fermat on Arithmetica by Diophantus.

It would be easier if you saw these famous diophantine equations before, rather than work it out thyself. Fermat sates that,
$x^4+y^4=z^2$
$x^4-y^4=z^2$
Has no solutions.
(The first one leads to Fermat's Last Theorem $n=4$.)

Consider a Pythagorean triangle where $z$ is hypotenuse. And $x,y$ are the legs.
Thus,
$x^2+y^2=z^2$. (1)
$A=\frac{1}{2}xy$.
Assume,
$\frac{1}{2}xy=w^2$.
Then,
$2xy=4w^2=(2w)^2$ (2)
Add (2) and then subtract (2) from (1),
$(x+y)^2=z^2+(2w)^2$ (3)
$(x-y)^2=z^2-(2w)^2$ (4)
Multiply (3) and (4) together,
$(x^2-y^2)^2=z^4-(2w)^4$
-----
Now CaptainBlank's question.

Let, $N$ be a positive integer.
And let,
$n_1,n_2,...,n_p$
Be all its factors (trivial and improper included).
Let,
$M=n_1n_2....n_p$ (*)
$M=n_1n_2....n_p$ (**)
Multiply them together,
But before we do let me explain what happens.
For each factor $n_i$ in (*) we can find one in (**) $n_j$ such that $n_in_j=N$.
Thus,
$M^2=N\cdot N\cdot ... \cdot N=N^p$
Thus,
$M=\sqrt{N^p}$.

(In the special case when $N$ is a square then $p$ is odd but $N$ is a square).