Prove that there is no primitive Pythagorean triple $(a,b,c)$ where one side differs from another by three I tried using all the Theorems I know about Pythagorean triples, but nothing is working out. I know I should probably go for a proof by contradiction.
 A: Use the parametric form:
$$a=r^2-s^2,$$
$$b=2rs,$$
$$c=r^2+s^2.$$
If say $a-b=3$ then $3=r^2-2rs-s^2=(r-s)^2-2s^2$. I think you can use congruences modulo some small number $n$ to dispose of this case.
Then you need to tackle $b-a=3$, $c-a=3$ and $c-b=3$.
A: A primitive Pyth. triplet is of the form $(a,b,c)=(m^2-n^2, 2mn, m^2+n^2)$ where $m>n>0$ and $\gcd (m.n)=1 ,$ and where $m,n$ are not both odd. (BTW: If $m,n$ were both odd then $a,b,c$ would  all be even.) We have $\gcd (a,b)=1.$
(1).  We have  $c-a=(m^2+n^2)-(m^2-n^2)=2n^2\neq 3.$
(2). We have $c-b=(m^2+n^2)-2mn=(m-n)^2\neq 3.$ 
(3A). If $m\equiv n \pmod 3$ then $n\not  \equiv 0\pmod 3,$ else $m=(m-n)+n\equiv 0\pmod 3,$ but then $3$ divides both $m$ and $n$,  contrary to $\gcd (m,n)=1.$ 
So if $m-n\equiv 0 \pmod 3$ then $n^2\equiv 1 \pmod 3,$ so modulo $3$ we have  $a-b\equiv (m^2-n^2)-2mn=(m-n)^2-2n^2\equiv 0-2 \not \equiv 0.$
(3B).   If $m-n \not \equiv 0 \pmod 3$ then $(m-n)^2\equiv 1 \pmod 3,$ so $a-b=(m-n)^2-2n^2\equiv 1-2n^2 \not \equiv 0\pmod 3.$ Because there is no solution to $1-2n^2\equiv 0 \pmod 3.$
Another way of handling cases (3A) and (3B): By contradiction:  If $\{a,b\}=\{x,x+3\}$ then  $x\equiv \pm 1 \pmod 3$ (Else $3$ divides both $x$ and $x+3$, that is, $3$ divides both $a$ and $b,$ contrary to  $\gcd (a,b)=1$).
But then $x^2\equiv 1 \pmod 3$, so modulo $3$ we have $c^2=a^2+b^2= x^2+(x+3)^2\equiv  2x^2\equiv 2.$ But there is no solution to $c^2\equiv 2 \pmod 3.$
