Triangles...?

Question

The lines of a right-angled triangle are positive integers: a, b, c. Prove that if the hypotenuse c is divided by 3 (without a remainder) than, each of the other two lines is divided by 3 (without a remainder).

Answer 1

Curt made it way too complicated

Since c is dived by 3, you can write c=3k (k is an integer)

Then a^2+b^2 = 9 k^2
Divide by 9 or 3^2

(a/3)^2 + (b/3)^2 = k^2, since k^2 is an integer, the only way for this expression to be an integer is for a and b to be divisible by 3

Answer 2

That's an algebra question, not geometry.

a^2 + b^2 = c^2.

For all A, either there exists some X such that A = 3X or there exists some X such that A = 3X + 1 or there exists some X such that A = 3X + 2.

Then the remainder of A^2 when divided by 3 equals either 1, 1, or 0.

The same is true of B^2.

So the remainder of (A^2 + B^2) when divided by 3 is either 0, 1, or 2.

But we're told it's 0.

What does that tell us about A and B?