What I Discovered on Right Angled Triangle?
Though Fermat did it earlier …
Motivation
This work was born as an outcome of a discussion with one of my savvy acquaintances. His challenge was : ‘Can you prove that there are or there are not two right angled triangles that share a leg and the hypotenuse of one is a leg at the other, where all the sides are integers”
Namely can we have a two right angled triangles that the first one has legs
a,b and hypotenuse c and the second has legs b, c and hypotenuse d. a, b, c, d are all integers
Since I believe in Pythagoras I can write his claim as follow:
Congruent Numbers
Before we move forward to my work I will introduce the concept of congruent numbers
Basically these are integers that can be an area of a right angled triangle with rational sides. One can formulate this by denoting a, b, c rational numbers and say that n is congruent if:
By doing some technical work we can say that n is a congruent number if for a, b, c, d rational numbers it satisfies
If we take the equations above
The question becomes “Is 1 a congruent number?”.
Somewhere in the past I heard it is not. But you know a challenge from a savvy acquaintance requires thoughts and not browsing.
Let’s Work
We denote the pair of equations by pair0. We assume that a, b and c do not have common factors . Since they are integers we can use rule of mode 4 to conclude that b is even
Strategy
We know that if pair of equation such as pair0 exist then there is an such pair with the smallest b. We will study this case aiming to exclude this b
We can write the upper equation
Since c, a are odd numbers we can conclude by mod(4) arithmetic that only one of them is a multiplication of 4 . Therefore we can define
We obtain:
We can deuce that
- For each pair (s, t) one of the numbers is odd (due to mod 4 arithmetic)
- t differs from t1
Thus s and s1 cannot be 1 simultaneously
=> b has an odd factor
We wish to exclude the case that t1is odd, and exclude the entire concept (Fermat did it already but I’ll go in my way). The next claim appears to be beneficial for this
Lets add an additional information to what we know on t1
We emphasize p, z>1 (since t1 is not a prime). Regarding r, w they are odd but can be 1.Let’s move gaps:
Hope you enjoyed
