Pythagorean Theorem and Triples

One of the most fundamental and groundbreaking theorems of the human history: Pythagoras’s Theorem

Pythagorean Theorem is called by the name of a Greek mathematician, Pythagoras.

Pythagorean Theorem’s Proofs

For a theorem like this, before examining its applications, one must definitely observe the proofs of this theorem and even try to prove it. Currently, there exists 367 proofs for this theorem! Of course, we won’t be examining all of them, however, let’s see how the two most known proofs look like.

First Proof: Let’s construct a square with the side of a+b. Then let’s draw 4 right triangles — sides’ lengths being a and b and the hypotenuse being c— on 4 corners as shown below.

We can express the area of the square in two different ways:

(a+b)²=4(ab/2)+c²

After basic algebraic simplification, we get the result: a²+b²=c².

Second Proof: We will use the power of a point in this proof. First, let’s construct a circle with radius length b and centre O. Take a point -let’s call it P- outside of the circle with |OP|=c and the length of the tangent which is drawn from P to the circle is a.

Pow(P)=a²=(c-b)(c+b)=c²-b² → a²+b²=c²

Pythagorean Triples

Now, let’s consider the equation a²+b²=c² where a, b and c is positive integers. How many solutions are there? Infinite. Substitute a=n²-1, b=2n and c=n²+1. It satisfies the Pythagorean equation and it shows that there are infinitely many solutions. However, there are more triples which satisfy the Pythagorean equation. For example (5,12,13) satisfies the equation but it is not in the form of (n²-1,2n,n²+1). We will find all form of triples.

We will show the greatest common divisors of a and b as (a,b) briefly.

If (a,b)=d then a=md, b=nd where (m,n)=1. a²+b²=d²(m²+n²)=c² → d divides c → c=kd where (m,n)=(n,k)=(k,m)=1 and m²+n²=k². So we reduced the equation to m²+n²=k².

Let’s find out triples which satisfy m²+n²=k².

Lemma 1: m and n cannot be both odd.

Proof of Lemma 1: If x is odd, then x²=(2k+1)² ≡ 1(mod 4). Hence m²+n²≡2(mod 4). However, a square can be 0 or 1 modulo 4. →←

Lemma 2: (k-n, k+n)=1

Proof of Lemma 2: By Euclid’s Algorithm 1=(n,k)=(n-k,k)=(n-k,2k)=(n-k,n+k). Since n and k are not both odd or even, we wrote (n-k,k)=(n-k,2k).

Applying these two Lemmas it follows

m²+n²=k² →m²=(k-n)(k+n), (x,y)=1, m²=x²y² →k+n=x², k-n=y²

Therefore, (m,n,k)=(xy,(x²-y²)/2,(x²+y²)/2)

Actually, for all integer x and y (2dxy,d(x²-y²),d(x²+y²)) is a Pythagorean Triple. Can you prove that all Pythgorean Triples can be written in form of (2dxy,d(x²-y²),d(x²+y²)).

References,

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