Number Theory – Are All Positive Integers Greater Than 2 Part of a Pythagorean Triple?

diophantine equationsdivisibilityelementary-number-theorynumber theorypythagorean triples

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 $2mn$.

Related Question