Pythagorean Triples : Is every positive integer $\gt$ $2$ part of at least one Pythagorean triple?

Question

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 ...

Answer 1

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 $2mn$.

Answer 2

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,

$$n^2 + (n-2)^2 + (2n+1)^2 + (3n^2+2)^2 = (3n^2+3)^2$$