Prove that if a b c is a Pythagorean triple with GCDa b c 1

Prove that if (a, b, c) is a Pythagorean triple with GCD(a, b, c) = 1, then GCD(a, b) = 1.

Solution

If possible let gcd (a,b) = u where u>1

then a= ku and b =lu for some integer k & l

As (a,b, c ) is Pythagorean triple, we may write a²+b²=c²  => c² = k²u² + l²u² = u²(k² +l²)

=> c = up where p is the the square root of (k² +l²)

So a= ku, b= lu , c = pu => gcd (a,b,c) = u or a multiple of u => it contradicts gcd(a,b,c) =1 since u>1

Thus our assumption that gcd(a,b) =u >1 is wrong .

therefore gcd(a,b)=1 (proved)