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The way I am planning to arrange this is by providing fragments of the proof, so I can understand what's going on before forging ahead, so if you are going to help me, keep in mind that I am going to be updating the thread with the missing fragments. I appreciate so much your help!.

Theorem 7-1. A general primitive solution of $x^2+y^2=z^2$, $y$ even, $x>0,y>0,z>0$,

is given by $$x=a^2-b^2,y=2ab,z=a^2+b^2$$

where $a$ and $b$ are prime to each other and not both odd, and $a>b>0$.

The above theorem and the proof below are from LeVeque's Elementary Theory of Numbers.

Proof: It is easily verified that for every such pair of integers a and b, the corresponding integers $x,y,z$ satisfy all the requirements. It remains to show that every solution arises from suitably chosen a and b satisfying the conditions of the theorem. (It's not clear to me what we are trying to prove and besides that, how do you verify for every such pair of integers a and b, the corresponding integers $x,y,z$ satisfy all the requirements).

Suppose that $x^2+y^2=z^2$. Since $(x,y,z)=1$, we also have $(y,z)=1$, so that $(z-y,z+y)=1$ or $2$. But z is odd and y is even and therefore $(z-y,z+y)=1$. (why is $(y,z)=1$?) I need a deeper explanation of the latter).

$$x^2=(z-y)(z+y)$$

We deduce that $z-y$ and $z+y$ must be odd squares (How?, an odd square is a number in the form $x^2$ that is odd right?), since they are positive. Now if $t$ and $u$ are integers of the same parity (both even or both odd), there are integers $a$ and $b$ such that $t=a+b$ and $u=a-b$ (why are we doing this?), namely $a=(t+u)/2$ and $b=(t-u)/2$ (ok here we just solve for b and a). Applying this in the case where $t$ and $u$ are the odd numbers of which $z+y$ and $z-y$ are the squares, respectively, we can set

$$z-y=(a-b)^2\\ z+y=(a+b)^2$$

(How is the latter true?)

Martin Sleziak
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First of, this fact is useful. (And I will use it below.)

Fact 1. For integers such that $x^2+y^2=z^2$ the following conditions are equivalent:

  • $\gcd(x,y,z)=1$
  • $\gcd(x,y)=1$
  • $\gcd(x,z)=1$
  • $\gcd(y,z)=1$

In other words, if two numbers in a Pythagorean triple are coprime, then it is a primitive Pythagorean triple.

I leave for you to think a bit about this. (Just think about what happens if some $d\mid x$ and $d\mid y$ for some $d>1$. Other cases are similar.)

Another useful fact:

Fact 2. If $m$ and $n$ are integers such that $\gcd(m,n)=1$ then $\gcd(m+n,m-n)$ is either $1$ or $2$.

See Prove $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ if $\gcd(a,b) = 1$

It's not clear to me what we are trying to prove and besides that, how do you verify for every such pair of integers a and b, the corresponding integers $x,y,z$ satisfy all the requirements.

Suppose $a>b>0$ are coprime and they have different parity. Then for $x=a^2-b^2$, $y=2ab$, $z=a^2+b^2$ we have.

  • $x^2+y^2=z^2$, which can be verified simply by plugging in and algebraic manipulation: $$(a^2-b^2)^2+(2ab)^2=a^4-2a^2b^2+b^4+4a^2b^2 = a^4+2a^2b^2+b^4=(a^2+b^2)^2$$
  • $y$ is clearly even and it is also clear that $x,y,z>0$;
  • $\gcd(x,z)=\gcd(a^2-b^2,a^2+b^2)$ can be only $1$ or $2$. But since $a$ and $b$ have different parity, neither of the numbers $x$ and $z$ is even. To we have $\gcd(x,z)=1$.

In summary: The above shows that the numbers $x$, $y$, $z$ indeed form a primitive Pythagorean triple.

Why is $(y,z)=1$?

See Fact 1 above.

...and therefore $(z-y,z+y)=1$. I need a deeper explanation of the latter

By Fact 2 we know that $\gcd(z-y,z+y)$ is either $1$ or $2$. But since both $z-y$ and $z+y$ are odd, it cannot be $2$. So the remaining possibility is $\gcd(z-y,z+y)=1$.

We deduce that $z-y$ and $z+y$ must be odd squares. (How?, an odd square is a number...)

If $x^2$ is a square and it is product of two coprime numbers, then each of them must be square, see: If a and b are relatively prime and ab is a square, then a and b are squares.

Similarly, each of the factors must be odd, if you want get an odd number.

... $z-y=(a-b)^2$, $z+y=(a+b)^2$. How is the latter true?

We know that $z+y=t^2$ and $z-y=s^2$ for some integers $s$ and $t$. We also know that $s$ and $t$ are both even. So there exists $a$, $b$ such that $a+b=t$ and $a-b=s$, meaning that \begin{align*} z+y&=(a+b)^2\\ z-y&=(a-b)^2 \end{align*}

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