Again Pythagoras

Number Theory Level 3

Consider the triplets of positive integers $(a,b,c)$ with $a$ is odd, $\gcd(a,b,c)=1$ and $a^2+b^2=c^2$ .

Which one can't be a value of $b$ ?

Notation: $\gcd(\cdot)$ denotes the greatest common divisor function.

1600 1006 6100 1060

1 solution

Brian Charlesworth
Dec 22, 2016

All primitive Pythagorean triples can be represented as $(m^{2} - n^{2}, 2mn, m^{2} + n^{2})$ for positive integers $m,n$ . Since $a$ is odd we must have $a = m^{2} - n^{2}$ and $b = 2mn$ . Now to ensure that $a$ is odd we must have one of $m,n$ be odd and the other even, which means that $b = 2mn$ must be divisible by $4$ . The only given option that is not divisible by $4$ is $1006$ , and hence $b$ cannot be $\boxed{1006}$ .

Again Pythagoras

1 solution

0 pending reports