# Form of even perfect numbers [closed]

From the list of even perfect numbers http://en.wikipedia.org/wiki/List_of_perfect_numbers it can be observed that all of them have either 6 or 8 as a last digit. Is this true for all even perfect numbers? In other words, does one of the congruences $$n\equiv 1 \ (\text{mod 5}), \quad n\equiv 3 \ (\text{mod 5})$$ hold for any even perfect number n? I suppose there are results of this kind but couldn't find any.

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## closed as off topic by Steven Landsburg, Gjergji Zaimi, Andrés E. Caicedo, Felipe Voloch, Noah SnyderJun 16 '12 at 15:36

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It follows from Euler's result that they are all of the form $2^{p-1}(2^p-1)$, where $2^p-1$ is a Mersenne prime. (In fact, all you need is for $p$ to be odd, or $2$ I guess.) – Henry Cohn Jun 16 '12 at 13:29

Every even perfect number is of the form $2^{p-1}(2^p - 1)$ where $2^p - 1$ is a (Mersenne) prime. Note that $p$ must be prime – if $p = ab$ with $a, b > 1$ then

$$2^p - 1 = 2^{ab} - 1 = (2^a)^b - 1 = (2^a - 1)(1 + 2^a + 2^{2a} + \dots\ + 2^{a(b-1)}).$$

If $p = 2$, we obtain the first perfect number $6$ which satisfies $6 \equiv 1\ (\text{mod 5})$. Every other prime is odd, so let $p = 2k + 1$. Then

$$2^{p-1}(2^p - 1) = 2^{2k}(2^{2k+1} - 1) = 2.2^{4k} - 2^{2k} = 2.16^k - 4^k \equiv 2 - (-1)^k\ (\text{mod 5}).$$

So, for $p = 2k + 1$,

$$2^{p-1}(2^p - 1) \equiv \begin{cases} 1 \ (\text{mod 5}) & \text{if }k\text{ is even}\newline 3 \ (\text{mod 5}) & \text{if }k\text{ is odd}. \end{cases}$$

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Thank you Michael, very nice proof. I have the following conjecture about twin primes and perfect numbers. There is never a perfect number between twin primes, with 6 the only exception. – Martin Jun 16 '12 at 14:20
Hi Martin. This is true because for any three consecutive numbers, exactly one of them is divisible by three. If $2^{p-1}(2^p - 1)$ is between a pair of twin primes, then it must be divisible by three. As three is prime and $2^{p-1}$ is not divisible by three, $2^p - 1$ must be divisible by three. But $2^p - 1$ is prime, so we must have $2^p - 1 = 3$. This only occurs when $p = 2$, which corresponds to the perfect number $6$. – Michael Albanese Jun 16 '12 at 14:42
Note, if a perfect number is between twin primes, it must be even and hence have the form mentioned above. – Michael Albanese Jun 17 '12 at 6:16