Existence of primitive Pythagorean triples

2017-12-14 05:22:41

Let $(a,b,c)$ be a Pythagorean triple, which means $c^2=a^2+b^2$.

If $c$ is odd and $a$ & $b$ are relatively prime, then there exist integers $m$ and $n$ such that $c=m^2+n^2, ~a=m^2-n^2, ~b=2mn$.

One can easily check the above by proving $\gcd(\frac{a+c}{2},\frac{a-c}{2})=1$.

My question is whether the converse of the above also holds or not.

;Whenever the odd number $c$ is a sum of 2 squares, then there exist integers $a$ and $b$ satisfying $c^2=a^2+b^2$ and $\gcd(a,b)=1$.

Any help will be appreciated.

The odd number $45$ is a sum of two squares, but not of two relatively prime squares.

More generally, let the odd number $n$ be of the form $p_1^{2a_1}\cdots p_k^{2a_k}m$, where the $p_i$ are primes congruent to $3$ modulo $4$, and $m$ is a product of primes congruent to $1$ modulo $4$. Then $n$ is a sum of two squares, but is not the sum of two relatively prime squares.

  • The odd number $45$ is a sum of two squares, but not of two relatively prime squares.

    More generally, let the odd number $n$ be of the form $p_1^{2a_1}\cdots p_k^{2a_k}m$, where the $p_i$ are primes congruent to $3$ modulo $4$, and $m$ is a product of primes congruent to $1$ modulo $4$. Then $n$ is a sum of two squares, but is not the sum of two relatively prime squares.

    2017-12-14 06:59:33