If $n$ is a perfect square number then $\sigma(n)$ is odd number. How to prove that if $n$ is a perfect square number then $\sigma(n)$ is odd number.
This $\sigma(n)$ is the sum of all divisors of $n$.
 A: If $p$ is a prime, then
$\sigma(p^{2n}) = 1+p+p^2+\cdots +p^{2n}$ is a sum of $1$ and $2n$ numbers of the same parity and so is odd.
Since $\sigma$ is multiplicative,
$\sigma(p_1^{2n_1} \cdots p_m^{2n_m})=\sigma(p_1^{2n_1}) \cdots \sigma(p_m^{2n_m})$ is the product of odd numbers and so is odd.
A: $$n=p_1^{\alpha}*p_2^{\beta}*p_3^{\gamma}...\\\sigma(n)=\frac{p_1^{\alpha+1}-1}{p_1-1}*\frac{p_2^{\beta+1}-1}{p_2-1}\frac{p_3^{\gamma+1}-1}{p_3-1}=\\(1+p_1+p_1^2+p_1^3+...+p_1^{\alpha})(1+p_2+p_2^2+...+p_2^{\beta})(1+p_3+p_3^2+...+p_3^{\gamma})\\ $$if $n$ is a perfect square 
it is inform of $$ p_1^{2\alpha}*p_2^{2\beta}*p_3^{2\gamma}...\\so\\\sigma(n)=(1+p_1+p_1^2+p_1^3+...+p_1^{2\alpha})(1+p_2+p_2^2+...+p_2^{2\beta})(1+p_3+p_3^2+...+p_3^{2\gamma})$$ multiplication of some odd number =odd number 
because $$(1+p_1+p_1^2+p_1^3+...+p_1^{2\alpha})=1+2\alpha (number)=odd$$ if $$p_1=odd \\p_1^n=odd\\(1+p_1+p_1^2+p_1^3+...+p_1^{2\alpha})=\\1+odd+odd+...+odd=1+2\alpha(odd)=\\1+even=odd$$
  if $$p_1=even \\ p_1^n=even\\(1+p_1+p_1^2+p_1^3+...+p_1^{2\alpha})=\\1+even+even+...+even=1+2\alpha (even)=\\1+even=odd $$
