Pythogorean Triplets

i am sorry have to post more of the question .. please wait

2) take this as the second question

Well the question i wanted to ask was that if x,y,z form a pythogorean triplet and they are a primitive triplet that is a triplet in the simplest form

this means 8,6,10 can be simplified back to 3,4,5 so primitive triplet means gcd(x,y,z)=1 then u have to prove that x,y,z are pair wise co prime

2 one is easy...pls give some time for the 1st one
Let x,y,z be a primitive soln, by definition gcd(x,y,z) = 1 cz otherwise it wont be the simplest.

then if a prime divide any two of x,y,z it must divide the other..and that explains why the numbers must be pair wise coprime

This has a quick answer if you use the theory of rings.

We write the equation as p=(x+iy)(x-iy)

The numbers of the form a+bi, a, b \in \mathbb{Z} are called Gaussian integers and they form what is known as a UFD (Unique Factorization Domain), i.e. just like integers they can uniquely be resolved into prime factors (up to units which here are ±1,±i)

Now we claim that x+iy is a prime.

Otherwise we would be able to write x+iy = (a+bi)(c+di)

where a^2+b^2 \ne 1; c^2+d^2 \ne 1

But that would mean p = (a^2+b^2)(c^2+d^2) which is absurd.

Likewise x-iy is also a prime

So in the ring of gaussian integers p = (x+iy)(x-iy) is the only factorisation possible.

It follows that there is up to permutation and sign, a unique solution to x²+y²=p

i guess there is a solution with more elementary concepts, which is what b555 may be looking for. So please do try

honestly i coudnt think of a simpler proof which is why i made use of "heavy machinery" :D.

Let p = x^2+ y^2 = m^2 + n^2

then p^2= (mx+ ny)^2+ (xn - ym)^2 (*), p^2= (nx+ my)^2+ (mx- yn)^2

Now comes my trick [3]

p|mn(x^2 + y^2)+ xy(m^2+ n^2)\Rightarrow p|(xm+yn)(xn+ym) [4]

Routine finishing left.

It follows that

Case 1: p|(xm+yn)\Leftrightarrow p^2|(xm+yn)^2\Rightarrow p^2\leq (xm+yn)^2\Rightarrow (xn-ym)^2= 0(From (*)) \Rightarrow xn= ym

Let \frac{x}{y}= \frac{m}{n}= k\neq 0

We had x^2+ y^2 = m^2+n^2\Leftrightarrow y^2(1+k^2) = n^2(1+k^2)\Leftrightarrow y^2=n^2

Case 2: p|xn+ym. Same way as above, we can show that y^2= m^2

had to work since it was posted..:D