Show that $n = 2^{m - 1} (2^{m} - 1)$ is a perfect number, if $(2^{m} - 1)$ is a prime number.

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Solution

As we know by the definition of perfect numbers that if the sum of the divisors of a number n, then n is called a perfect number.Let $n = 2^{m - q} \times p$ where $p = 2^{m} - 1$ is prime number.Divisor of $2^{m - 1} \times p$ are $1, 2, 2^{2}, 2^{3}, . . . . . . {.2}^{m - 1}, p, 2 p, 2^{2} p, . . . ., 2^{m - 2} p, 2^{m - 1} p .$
Now we should sum all these divisors except the last one, i,e., $2^{m - 1} p$
$S = (1 + 2 + 2^{2} + . . . . + 2^{m - 1}) + p (1 + 2 + 2^{2} + . . . . . . + 2^{m - 2})$
$= \frac{1 (2^{m} - 1)}{2 - 1} + \frac{p [1 (2^{m - 1} - 1)]}{2 - 1}$
$= 2^{m} - 1 + p (2^{m - 1} - 1)$
$= 2^{m} + p m^{m - 1} - p - 1$
$= 2^{m - 1} (2 + p) - (p + 1)$
$= 2^{m - 1} (1 + 2^{m}) - 2^{m} [s i n c e, p = 2^{m} - 1]$
$= 2^{m - 1} (2^{m} - 1) = n$

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