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 …
Best Answer
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 $2mn$.