Too easy for TST (did it "by typing") (a) Write $N=p_1p_2p_3...p_k$ where $p_i$ are distinct primes. To be a perfect number we need $2p_1p_2...p_k=(p_1+1)(p_2+1)\cdots (p_k+1)$ since RHS is the sum of all positive divisors of $N$. If $p_i$ are all odd then we can have at most 1 prime divisor, else the power of 2 in RHS exceeds that in LHS, which is just 1. Therefore, for the odd case we have $2p_1=(p_1+1)$ or $p_1=1$, contradiction. Otherwise, Wlog $p_1$ is even, so $p_1=2$. Thus, we have $4p_2p_3...p_k=3(p_2+1)(p_3+1)\cdots (p_k+1)$. By applying a similar argument to the first case, we see that if $k\ge 4$ we get a contradiction since the power of 2 in LHS is 2 and that of RHS would be at least 3. Therefore, $k\le 3$. For $k=3$ we get $4p_2p_3=3(p_2+1)(p_3+1)$ which leads to $(p_2-1)(p_3-1)=10$. But the exponent of 2 in LHS is at least 2 and that in RHS is 1, contradiction. For $k=2$ we get $4p_2=3(p_2+1)$ or $p_2=3$, which works. Thus, $N=p_1p_2=2\cdot 3=\bf 6$ is the only solution. Indeed, $2\cdot 6=1+2+3+6$. (b) Suppose $N=p_1^{2e_1}p_2^{2e_2}\cdots p_k^{2e_k}$ is a perfect number. Assume for contradiction that $N$ is perfect, i.e. $2p_1^{2e_1}p_2^{2e_2}\cdots p_k^{2e_k}=\prod_{i=1}^k (1+p_i+p_i^2+\cdots +p_i^{2e_i})$. Now if all the $p_i$ are odd then since each each bracket in RHS is the sum of $2e_i+1$ odd numbers, RHS is the product of odd numbers and thus odd. But LHS is even, contradiction. On the other hand, suppose Wlog $p_1=2$. Then $2^{2e_1+1}p_2^{2e_2}p_3^{2e_3}\cdots p_k^{2e_k}=(1+2+2^2+\cdots +2^{2e_1})\prod_{i=2}^k (1+p_i+p_i^2+\cdots +p_i^{2e_i})$. As before, all the terms on RHS except for $(1+2+2^2+\cdots +2^{2e_1})$ are odd from the previous case, and we can show that this term is odd too by observing that it is the sum of $2e_1$ even numbers and an odd number, hence of the form $2k+1$, i.e. odd. But the exponent of 2 in LHS is $2e_1+1\ge 3$, contradiction.