Proof question?

Question

If anyone could explain how the following is done, it would be greatly appreciated!

Prove that in any primitive Pythagorean triple (a,b,c) either a or b must be a multiple of 3.

Answer 1

Yes, Tony (answerer 2) has it.

The square of an integer is either a multiple of 3, when the integer is a multiple of 3, or one more than a multiple of 3, when it isn't. The square simply cannot be two more than a multiple of 3. We only need to list the three possibilities 0^2, 1^2 and 2^2 to prove this, knowing that adding multiples of 3 before squaring will only add multiples of 3 to the square.

So if neither a nor b is a multiple of 3, then the left-hand side of a^2 + b^2 = c^2 is the sum of two numbers each of which is one more than a multiple of 3, therefore comes to two more than a multiple of 3, which cannot be equal to the right-hand side no matter what, therefore the starting assumption is false. The opposite of the starting assumption is that at least one of a and b is a multiple of 3, so that must be true.

Answer 2

Consider congruences (mod 3). Assume that neither a nor b is 0 (mod 3). Then either a and b are 1 or 2 (mod 3); but then both a^2 and b^2 are 1 (mod 3). But then c^2 = a^2 + b^2 = 2 (mod 3), and c^2 = 2 (mod 3) must offend the sensibilities of the cognoscienti because squares cannot be 2 (mod 3). Ergo, you got yourself a contradiction, Chollie.

Answer 3

a = x^2 - y^2, b = 2xy, c = x^2 + y^2,
and gcd(a,b,c)=1

just study the cases 3k, 3k+1, 3k+2 for a and b and c