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. 

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.
