# Sum of two perfect squares is also a perfect square. Proof that one of these numbers is divisible by 3

At first, I tried to proof by contradiction: I considered two numbers that are not divisible by $$3$$. Than I tried to write consecutive perfect squares that are divisible by $$3$$ to see some pastern, but it didn't get me anywhere. Then i tried to make express each whole number a,b and c under the condition that $$a^2+b^2=c^2$$ in terms of two arbitrary integers. I found out that $$a$$ has to be of the form $$2mn$$ and $$b$$ has to be of the form $$m^2-n^2$$, where $$m$$ and $$n$$ are whole numbers. So that I have to prove that either $$2mn$$ or $$m^2-n^2$$ is divisible by $$3$$. I got stuck at this point and don't know how to prove that, so help me please

• $$x^2\equiv0\text{ or }1\mod{3}$$ Feb 1 '20 at 22:37

HINT. Write $$c^2=a^2+b^2$$ and consider the equation $$\mod 3$$, remembering that $$x \equiv 0 \mod 3$$ implies that $$x$$ is divisible by $$3$$.

Squares are $$0$$ or $$1$$ modulo $$3$$, so a sum of two of the latter can't be a square.

Your first strategy will turn out to be the more effective one.

Obviously, if a number is divisible by 3, then its square will be as well. Otherwise, the number is of the form $$3k\pm1$$ for some integer $$k$$, and $$(3k\pm1)^2=9k^2\pm6k+1\equiv1\pmod3$$

Therefore, the square of a number is either divisible by 3 or it leaves a remainder of 1.

Now, let us consider that there are positive integers $$a,b,c$$ such that $$a^2+b^2=c^2$$. From our earlier notion, it cannot be that neither $$a$$ nor $$b$$ are multiples of 3, because then their sum would be of the form $$3k+2$$ and could not be a perfect square. Therefore, either $$a$$ or $$b$$ (or both) are divisible by 3.

• Thanks for the answer, but i have two questions. Why the number of the form 3k+2 can't be a perfect square? And if only one of these numbers is divisible by 3 then the sum of their squares will be of the form 3k+1? Feb 1 '20 at 23:31
• Any number is $3k$ or $3k\pm1$, and none of those square to $3n+2$ Feb 2 '20 at 0:23