Prove that it does not exist two number such that $m^2+n^2=6 \underbrace {0 \cdots 0}$

Question

Prove that it does not exist a two number $m,n\in \mathbb N$ such that $m^2+n^2=6 \underbrace {0 \cdots 0}$, in solution he choose to divide by 9, but i do not know why that number, I know that remainder of $m^2 \equiv r_1(mod 9)$ and $n^2 \equiv r_2(mod 9)$ $r_1,r_2 \in \{0,7,1,4\}$ $m^2+n^2\equiv r_3(mod 9)$ $r3\in \{1,2,4,5,7,8\}$ but $10\equiv1 (mod 9)$ $10^{13} \ equiv 1 (mod9)$ $6\cdot 10^{13}\ equiv 6 (mod9)$ so this is not true, but they choose a number nine because it is easy to prove or there is something else?

Answer 1

$60\ldots 0 = 6 \times 10^n$ for some $n$. The point is that $10 \equiv 1 \mod 9$, so $6 \times 10^n \equiv 6 \mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 \mod 9$.

Answer 2

Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even. Prime factorizaton of $6\underbrace {0\cdots 0}$ is

$$6\underbrace {0\cdots 0}= {3}\times {2^n}\times {5^m}$$

Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6\underbrace {0\cdots 0}$ cannot be expressed as sum of two integers