We should take a moment to explain why,

1) If N is not prime, then 2^N − 1 is not prime either:

If N = a b then

(2^a − 1) × ( 2^0a + ^1a +^2a + …+^(b-1)a )

= 2^{a b} − 1 = 2^N − 1

2) if P=2^p − 1 is prime, then Q = 2^p−1P is perfect:

A number is perfect if it is the sum of all of its proper divisors. The divisors of Q are

1, 2, 2², …, 2^p−1 and P, 2P, 2²P, …, 2^p−1P = Q

Q is perfect if all of these divisors sum to 2Q (Since we’re summing in Q itself, as well as all of the proper divisors of Q)

But this sum is

(1+2+…+2^p−1)+ (1+2+…+2^p−1)P

= (1+2+…+2^p−1)(P+1)

=(2^p-1)(2^p)

= (P) (2 × 2^p−1) = 2Q

3) Now this is a little harder, but not too bad if you follow closely: If an even number N is perfect, then it is of this form!

So, let N be an even perfect number; in particular, then, N = 2ⁿ X for some n≥1 and odd number X.

The divisors of N are all the numbers 2ⁱ x where 0 ≤ i ≤ n and x is a divisor of X .

If we let S be the sum of the divisors of X, then the sum of all of the divisors of N is thus

(2ⁿ⁺¹ − 1) S = 2N

(recalling that N is perfect!)

So

(2ⁿ⁺¹ − 1) S = 2ⁿ⁺¹ X

Since (2ⁿ⁺¹ − 1) is odd, we must have that 2ⁿ⁺¹ divides into S , and so for some q ,

S = 2ⁿ⁺¹ q

and

(2ⁿ⁺¹ − 1) 2ⁿ⁺¹ q = 2ⁿ⁺¹ X

Canceling from both sides we now have

X = (2ⁿ⁺¹ − 1) q

In particular, X is a proper multiple of q . Now add q to both sides of this last equation; we obtain

X + q = 2ⁿ⁺¹ q = S

Think about this: S is the sum of all of the divisors of X. If q ≠ 1, then there are at least three divisors of X, namely 1, q and X itself. But then we have a contradiction, since S would then be greater than X + q.

So: q = 1 and X = 2ⁿ⁺¹ − 1

and our original N is of the correct form!

4) BUT no one knows if there is an odd perfect number, or even if there are infinitely many even perfect numbers!