WakeUp wrote: Suppose that the sum of all positive divisors of a natural number $n$, $n$ excluded, plus the number of these divisors is equal to $n$. Prove that $n = 2m^2$ for some integer $m$. Let $n=2^k\prod p_i^{n_i}$ with all prime $p_i$ odd. The equation is $\left(\sum_{j=0}^{k}2^j\right)\left(\prod_i\left(\sum_{j=0}^{n_i}p_i^j\right)\right)-n+(k+1)\prod_i(n_i+1)-1=n$ And so $\left(\sum_{j=0}^{k}2^j\right)\left(\prod_i\left(\sum_{j=0}^{n_i}p_i^j\right)\right)+(k+1)\prod_i(n_i+1)=2n+1$ Looking at this equality modulus $2$, we get $\prod_i(n_i+1)+(k+1)\prod_i(n_i+1)\equiv 1\pmod 2$ And so $k\prod_i(n_i+1)\equiv 1\pmod 2$ So $k$ is odd and all $n_i$ are even. Hence the result.