Let $a,b,c$ form a primitive Pythagorean triple, with $a$ odd. Prove that $(c-b)$ and $(c+b)$ share no common prime factors. Let $a,b,c$ form a primitive Pythagorean triple (meaning $(a,b,c) \in \mathbb{Z}^{3}$). Let $a$ be odd. Use a proof by contradiction to show that $(c-b)$ and $(c+b)$ share no common prime factors.
My Thoughts
So I began by assuming that $(c-b)$ and $(c+b)$ shares at least one common factor.
And so, for some prime $p$, $(c-b) = xp$ and $(c+b) = yp$, such that $(x,y) \in \mathbb{Z}^{2}$
So now, I thought maybe if we set this equal to either $a^2$ or $c^2 - b^2$ I could find a contradiction. But in either case, I couldn't come to any conclusion. 
 A: You were on the right track.

From $c-b = xp\;$and $c+b = yp$, you get
$$a^2 = c^2 - b^2 = (c+b)(c-b) = xyp^2$$
hence $p^2{\,\mid\,}a^2$, so $p{\,\mid\,}a$.

Since $a$ is odd, and $p{\,\mid\,}a$, it follows that $p$ is odd.

From $p{\,\mid\,}(c+b)$ and $p{\,\mid\,}(c-b)$, we get $p{\,\mid\,}\bigl((c+b)-(c-b)\bigr)$ and $p{\,\mid\,}\bigl((c+b)+(c-b)\bigr)$.

Thus, $p{\,\mid\,}2b$ and $p{\,\mid\,}2c$, but then, since $p$ is odd, it follows that $p{\,\mid\,}b$ and $p{\,\mid\,}c$.

Then $p$ is a common factor of $a,b,c$, contrary to the assumption that the triple $(a,b,c)$ is primitive.
A: So now, $xp\cdot yp = (c+b)(c-b)=c^2-b^2 = a^2$ which is odd (since $a$ is given as odd). 
Thus $(c+b)$ and $(c-b)$ are both odd also. as are $x,y,p$. And $c = (xp+yp)/2 = p(x+y)/2$ is divisible by $p$, as is $a$, and thus $p$ must divide $b$ also, which contradicts the primitive nature of the triple.
A: 
And so, for some prime p, (c−b)=xp and (c+b)=yp, such that (x,y)∈Z2
So now, I thought maybe if we set this equal to either a2
  or c2−b2 I could find a contradiction. But in either case, I couldn't come to any conclusion

If your first idea was to let $p$ be prime so that $x*p = c-b$ and $y*p = c
=b$, then
Your first trick should be to note is $a^2 + b^2 = c^2 \implies a^2 = c^2 - b^2 = (c+b)(c-b)=xyp^2$ so $p|a$
The second trick is note that $p|(c-b)$ and $c|c+b$ means $p|(c-b) + (c+b) = 2c$ and $p|(c+b)-(c-b) = 2b$.  So if $p\ne 2$  then $p$ divides all of $a,b,$ and $c$.
And then the last idea is to note that the problem did state that $a$ was odd (so $p$ can't equal $2$), and that $a,b,c,$ were primitive (so no such prime can exist).  
(Although it could if $a$ were even and $b$ and $c$ were odd and $p=2$$\ ^*$. Or, of course, if $a,b,c$ weren't primitive and therefore any common divisor would have to also be a common divisor of $c-b$ and $c+b$.)
Those tricks will become more and more apparent and familiar with experience.
$\ ^*$ If $a$ is even then $2$is the only factor $c+b$ and $c-b$ will have in common.  $b$ and $c$ will both have to be odd (if one were even so would the other) and so $c+b$ and $c-b$ are both even.  One of $c+b$ or $c-b$ is divisible by $4$ but the other will not be.
A: Another way:
Using https://en.m.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple,
let $a=m^2-n^2,b=2mn, c=m^2+n^2$ where $(m,n)=1$ and $m,n$ have opposite parity
Now if  prime $p$ divides both $c\pm b=(m\pm  n)^2,$
$p$ must be odd and must divide  $m\pm n$ 
and hence will divide $m+n\pm(m-n)$ 
and hence $2(m,n)$
A: Since $a$ is odd, $b$ is even . Because if $a=2a'+1$ is odd and if $b=2b'+1$ is also odd then $c^2=(2a'+1)^2+(2b'+1)^2$ has a remainder of $2$ modulo $4,$ which is impossible for a square.
So $c^2$ is odd, so $c$ is odd . So $c+b$ and $c-b$ are both odd.
Let $d$ be a common divisor of $c+b$ and $c-b$. Note that $d$ must be odd.
Then $d$ divides  $(c+b)+(c-b)=2c$ and $d$ divides $(c+b)-(c-b)=2b.$ But  $d$  is odd, so $d$ must divide $c$ and $b .$
Since $d$ divides $c+b$ and $c-b$ it divides the product $(c+b)(c-b)=a^2.$  But if $d$ is  prime and $d$ divides $a^2$ then $d$ divides $a.$ This would make $d$ a common prime divisor of $a,b,c,$  which would make $(a,b,c)$  non-primitive.
A: I developed a formula that generates only and all triples where $(C-B)=(2n-1)^2,n\in\mathbb{N}$. This subset includes all primitives and none of the trivials, the doubles and the square multiples of primitives that Euclid's formula generates.
$$A=(2n-1)^2+2(2n-1)k\quad =2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2$$
This generates only primitives except when $k$ is multiple of any factor of $(2n-1)$. For this discussion,   $(2n-1)$ does not divied $k$ because we require a primitive triple.
We can see by definition that $$\frac{C+B}{C-B}=\frac{\big(2(2n-1)k+2k^2\big)+\big(2n-1)^2+4(2n-1)k+4k^2\big)}{(2n-1)^2} =\frac{(2n-1)^2+4(2n-1)k+4k^2}{(2n-1)^2}$$
The denominator shares factors with the first two terms of the numerator   but not with the third term of the numerator so $(C-B)$ and $(C+B)$ have no common factors.
