Pythagorean Triples : Is every positive integer $\gt$ $2$ part of at least one Pythagorean triple? I was doing some basic number theory problems from Rosen and came across this problem: 

Show that every positive integer $\gt$ $2$ is part of at least one Pythagorean triple

My Solution (partial) :
Case - 1 :

  
*
  
*Let there be an integer $t$ $\ge$ 3
  
*Suppose $t$ is of the form $2^{j}$ for $j > 1$
  
*Let $m$ = $2^{j-1}$ and $n$ = $1$
  
*So , $2mn$ = $t$ and hence $t$ belongs to a Pythagorean triple
  

Case - 2 :

  
*
  
*Let $t$ = $2n + 1$ 
  
*WLOG , let $m = n + 1$ 
  
*Then $m$ and $n$ have opposite parity
  
*Also , $m > n$ 
  
*So , $m^{2}$ - $n^{2}$ $=$ $2n + 1$ $=$ $t$, so $t$ belongs to a Pythagorean triple  
  

My Problem:
Can someone help me out ? I do not know if I am correct , I am all thumbs ; even a hint would suffice ...
 A: Using the characterisation of these triples, it suffices to show that any such number can be written as $m^2-n^2$, $2mn$ or $m^2+n^2$ with some numbers $m>n$.
The case $m^2-n^2$ covers "the most" numbers (only those $\equiv 2 \mod 4$ remain), the rest is covered by $2mn$.
A: 
I. Yes. Proof without words:

$$(\color{brown}{2m})^2+(m^2-1)^2 = (m^2+1)^2$$
$$(\color{brown}{2m+1})^2+(2m^2+2m)^2 = (2m^2+2m+1)^2$$

II. Higher. 

To prove it for quadruples is easier since even and odd cases can be combined into a single identity,
$$n^2+(n+1)^2+(n^2+n)^2 = (n^2+n+1)^2$$
and for quintuples,
$$n^2 + (n-2)^2 + (2n+1)^2 + (3n^2+2)^2 = (3n^2+3)^2$$
A: If $n$ is an odd integer, let $m = \frac{n^2 - 1}2$, then $m+1$, $m$ and $n$ are a Pythagorean triple. ($n^2 = 2m+1$)
If $n$ is even, let $m = \frac{n^2 - 4}4$, then $m+2$, $m$, and $n$ are a Pythagorean triple. $(n^2 = 4m+4)$
A: If $t$ is odd and $t\geq 3$ then  $m=(t+1)/2$ and $n=(t-1)/2$ are positive integers with $m^2-n^2=t.$ And $(m^2-n^2,2mn, m^2+n^2)=(t,2mn, m^2+n^2)$ is a P. triple.
If $t$ is even and $t\geq 4$ let $m= t/2$ and $n=1$. Then $m,n$ are positive integers with $m>n$ so $(m^2-n^2,2mn,m^2+n^2)=(m^2-n^2,t,m^2+n^2)$ is a P. triple. 
A: For primitive triples, side $A$ can be any odd number $>2$ and side $B$ can be any integer multiple of $4$. If we include multiples like $(6,8,10)$, then the even numbers that are not multiples of $4$ are include. Therefore, every $n>2\land n\in\mathbb{N}$ is part of at least one Pythagorean triple.
