How do you find out perfect numbers?

Answer 1

First let us define a perfect number. We call a number perfect if it is a positive integer and is the sum of all it proper divisors. The proper divisors part means we do not include the number itself even though any number divides itself. So for example 6 is perfect because it is a positive integer and 1+2+3=6. Not we excluded the number 6 even though 6 does divide 6.

Now we could look at any number and find all the divisors of the number except the number itself. Next add them all and if the sum is equal to the number itself. If so then that number is a perfect number. While this method will always work, it is quite tedious.

Euclid found another way thousands of years ago. He discovered that you could find the first four perfect numbers by using the formula 2p-1(2p − 1) where p is a prime such as 2,3,5 or 7. So let us consider the first prime 2 and plug it in Euclid's formula. 22-1 (22 -1)

is equal to 21 (4-1)=2x3=6 which is the first perfect number. It is important to notice that the (2p − 1) portion is always giving us a Prime number. The perfect number we would obtain with the primes 2,3,5, and 7 are also even. For example 23 -1=7 which is prime and we multiply that by 22 so 4x7=28 which is perfect and even.

Everything was great until we tried the 5th prime which is the number 11. When we plug that into the (2p − 1) part just as we did 2 and 3, we get the number 211 − 1 = 2047 = 23 × 89 so it is not a prime number like the values we get with 2,3,5, and 7. Why do we care about the fact the this is not prime? We were looking for perfect numbers not primes?

Because when we multiply 2047 by 211-1 we find out the the number we get is not perfect.

So Euler was correct only for those first few values.

When (2p − 1) is prime it has a special name. It is called a Mersenne prime named after Marin Mersenne who was a monk in the 17th ( a prime number by the way) century who studied math and primes and a field known as number theory. In order for (2p − 1) to be prime is has been found that it is necessary but not sufficient for p to be prime. In math a necessary condition is one that must hold true for the statement to be true. So p must be a prime. However a sufficient condition is one that says if this is satisfied the statement is always true. So the prime 11 showed us the (2p − 1) is not always prime and this means the condition is not sufficient. However, if p is not a prime (2p − 1) will not be prime either.

A century later, Euler proved that all perfect numbers are generated by (2p-1 )(2p − 1).

Euclid has tried but could not prove it. We can find the first 39 perfect numbers by using the primes: 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917

Note that all these number are even. We do not know if there are any odd perfect numbers. It has been shown that there are no odd perfect numbers in the interval from 1 to

1050 but we can't say in general.

We don't even know how many perfect numbers or Mersenne primes there are. While we suspect the number is infinite, that has not been proven yet.

So here are the first few perfect numbers and you will see they become quite larger very fast!

6,

28,

496,

8128,

33550336,

8589869056,

137438691328,

2305843008139952128,

2658455991569831744654692615953842176,

191561942608236107294793378084303638130997321548169216,

131640364585696483372397534604587229102234723183869431

17783728128,

144740111546645244279463731260859884815736774914748358

89066354349131199152128,

2356272345726734706578954899670990498847754785839260071014302

7597506337283178622239730365539602600561360255566462503270175

0528925780432155433824984287771524270103944969186640286445341

2803383143979023683862403317143592235664321970310172071316352

7487298747400647801939587165936401087419375649057918549492160

555646976,

1410537837067120690632079580860631898814867435147156678388386

7599995486774265238011410419332903769025156195056870982932716

4087724366370087116731268159313652487450652439805877296207297

4467232951666582288469268077866528701889208678794514783645693

1392206037069506473607357237869517647305526682625328488638371

5072974324463835300053138429460296575143368065570759537328128,

5416252628436584741265446537439131614085649053903169578460392

0818387206994158534859198999921056719921919057390080263646159

2800138276054397462627889030573034455058270283951394752077690

4492443149486172943511312628083790493046274068171796046586734

8720992572190569465545299629919823431031092624244463547789635

4414813917198164416055867880921478866773213987566616247145517

2696430221755428178425481731961195165985555357393778892340514

6222324506715979193757372820860878214322052227584537552897476

2561793951766244263144803134469350852036575847982475360211728

8040378304860287362125931378999490033667394150374722496698402

8240806042108690077670395259231894666273615212775603535764707

9522501738583051710286030212348966478513639499289049732921451

07505979911456221519899345764984291328

You can find every know Mersenne prime and hence every know perfect number on a site called GIMPS. I have included a link. But even more than seeing them, you can help discover the next one and become quite famous!

Now this nice answer on perfect numbers will conclude with an important theorem.

If 2k-1 is a prime number, then 2k-1(2k-1) is a perfect number and every even perfect number has this form. I will include a proof by Chris Caldwell. It uses the sigma function which is a function that finds the sum of the divisors. I will give link for the sigma function for those who are interested.

Proof: Suppose first that p = 2k-1 is a prime number, and set n = 2k-1(2k -1). To show n is perfect we need only show sigma(n) = 2n. Since sigma is multiplicative and sigma(p) = p+1 = 2k, we know

sigma(n) = sigma(2k-1)(sigma(p)) = (2k-1)2k = 2n.

This shows that n is a perfect number.

On the other hand, suppose n is any even perfect number and write n as 2k-1m where m is an odd integer and k>2. Again sigma is multiplicative so

sigma(2k-1m) = sigma(2k-1)(sigma(m)) = (2k-1)(sigma(m)).

Since n is perfect we also know that

sigma(n) = 2n = 2km.

Together these two criteria give

2km = (2k-1)(sigma(m)),

so 2k-1 divides 2km hence 2k-1 divides m, say m = (2k-1)M. Now substitute this back into the equation above and divide by 2k-1to get 2kM = sigma(m). Since m and M are both divisors of m we know that

2kM = sigma(m) > m + M = 2kM,

so sigma(m) = m + M. This means that m is prime and its only two divisors are itself (m) and one (M). Thus m = 2k-1 is a prime and we have prove that the number n has the prescribed form.

It is even easier to prove that if for some positive integer n, 2n-1 is prime, then so is n.

So this question about perfect numbers has generated some interesting discussion of Mersenne primes and included some exciting number theory. This is a beautiful topic in math that is often not covered.