The search for answers to simple questions like “why is the sky blue? ” and “why are there selfish acts” is a nice way to bite into a fair amount of science and philosophy, respectively. In math the same is true. Why do even-perfect numbers only end in 6 or 8? If you’re fortunate enough to have a child asking you such a question , all you need is about 1200 words to answer it—only 1/9 th of the number of words in a sitcom script. 🙂 And the discussion of perfect numbers begins with a venture into a special type of prime number.

Mersenne, a priest, was a mathematician but primarily a music theorist and a hub for scientific communication in the 1600s.

As of February 2022, of all the known primes, only 51 of them are Mersenne primes. A Mersenne prime *M*_{n} can be expressed as *M*_{n} = 2^{p} − 1, where *p* itself is prime. The largest prime to date is 2^{82 589 933} − 1, known as M51. To put things into perspective, our universe only has 10^{80} atoms, a number with only 81 digits. But M51 has 24 862 048 digits, meaning that it is bigger than the inconceivably large 10^{ 107}. Given that the number of primes less than 10^{ 107 }is approximately equal to 10^{ 107}/log(10^{ 107}), in comparison, 49 (plus the possibly few smaller Mersennes yet to be discovered), divided by that gargantuan number is an unimaginably small fraction indeed.

**Why ***p* must be prime if 2^{p} − 1 is prime

An interesting matter is why *p* must be prime if 2^{p} − 1 is prime. To convince ourselves, we will use a proof by contradiction—in other words we will make an initial assumption that states the opposite condition, logically follow it to a contradiction, and in the process reveal the assumption to be invalid.

We know from a geometric series that :

x^{n} − 1 = (x − 1)(x^{n − 1 }+ x^{n − 2} + · · · + a + 1)

For example x^{3} – 1 = (x-1) ( x² + x + 1)

Let’s assume that the exponent was composite, such that the Mersenne prime would be expressed as 2^{an} – 1. In other words we assume its exponent to be composite. Let x = 2ª. from the above geometric series; then

2^{an} − 1 = (2ª − 1)(2ª^{(n − 1)}+ 2ª^{(n − 2)} + · · · + 2ª + 1)

But now we have expressed a prime number as two factors, neither of which is equal to one, which is impossible. The original assumption about *p* being composite was therefore false, so **p** must be prime.

2. ** Getting a perfect number from a Mersenne Prime**

Next we will multiply the Mersenne prime by a power of 2 whose exponent is one less than the prime used to generate the Mersenne prime. In other words:

(2^{p-1 })(2^{p} − 1).

Interestingly this resulting number will always be a *perfect number*, a number that is the sum of all of its factors except itself. For example 28 is perfect because ( 2^{3-1 })(2^{3} − 1) = 1 + 2+ 4+7 + 14 = 28.

To reveal the “perfection” of any number of the form (2^{p} − 1) ( 2^{p-1 }) where p is prime, we remind you of the function σ(x). σ(x) is equal to the sum of all the factors of x *including x*. Adding a perfect number, N= ( 2^{p-1 })(2^{p} − 1), to a sum that already equals N, implies that in order for a number to be perfect,

σ(N) = 2N = 2 ( 2^{p-1 })(2^{p} − 1)= ( 2^{p })(2^{p} − 1) … (equation 1)

We start with the theorem σ(ab) = σ(a)σ(b), where a and b’s greatest common divisor =1. The reason this theorem is valid is that multiplying the sum of factors of two individual factors a and b is equivalent to getting all the *non-redundant* combinations of the product’s factors. For example, consider ab = 12 where σ(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28. That is equivalent to doing the following : σ(4)σ(3) = ( 1 + 2 + 4)(1 + 3) = 1(1) + 1(2) + 1(3) +2(3) + 4(1) +4(3) = 28.

Applying the theorem to our expression:

σ[( 2^{p-1 })(2^{p} − 1)] = **σ( 2**^{p-1 }) ∗ σ(2^{p} − 1). … equation (2)

Beginning with **σ( 2**^{p-1 }), we have here simply the sum of the factors of a power of two up to that power. **σ( 2**^{p-1 }) = 2^{0} + 2^{1} + 2^{2} + 2^{3} + …..2^{p-2} + 2^{p-1}.

We add 1 to each side:

**1+ σ( 2**^{p-1 }) =( 1 + 2^{0} )+ 2^{1} + 2^{2} + 2^{3} + …..2^{p-2} + 2^{p-1}.

We notice that the first terms’ sum is equal to the next power such that:

**1+ σ( 2**^{p-1 }) = (2^{1} + 2^{1} )+ 2^{2} + 2^{3} + …..2^{p-2} + 2^{p-1}. But this keeps happening:

**1+ σ( 2**^{p-1 }) = (2^{2} + 2^{2} )+ 2^{3} + …..2^{p-2} + 2^{p-1}. When we get to the second last term, we will have

**1+ σ( 2**^{p-1 }) = (**2**^{p-2} +2^{p-2} )+ 2^{p-1}.

**1+ σ( 2**^{p-1 }) = (2^{p-1} )+ 2^{p-1 }= 2^{p} ,

or **σ( 2**^{p-1 }) = 2^{p} – 1 ……equation (2b)

We substitute this result into equation (2) and obtain

σ[ ( 2^{p-1 })(2^{p} − 1)] = (**2**^{p} – 1 ) * σ(2^{p} − 1). … equation(3)

But σ(2^{p} − 1) is the sum of the factors of a prime number which means that we merely have to add 1 to itself = 1 + 2^{p} − 1 = **2**^{p} .

Substituting this result into equation (3) we obtain:

σ[( 2^{p-1 })(2^{p} − 1)] = (**2**^{p} – 1 )2^{p} , which meets the condition we had outlined for a perfect number in equation (1)!

**3. Why multiply a Mersenne by ( 2**^{p-1 })?

A derivation of S = n(n+1)/2

But where does the idea of multiplying a Mersenne prime (2^{p} − 1) by ( 2^{p-1 }) come from? To get the sum, S, of all the natural numbers from 1 to 100 in other words (1 + 2+ 3+ 4 ….100), we simply multiply the largest number by the next natural number and divide by 2, so S(100) = 100*101/2 = 5050. The formula is S(n)= n(n+1)/2 and I derived it informally and partly pictorially in the adjacent drawing.

Applying the formula to the Mersenne prime we get (2^{p} − 1)(2^{p} − 1 + 1)/2 = (2^{p} − 1)(2^{p})/2¹ = (2^{p} − 1)(2^{p-1})

**4. But does an even perfect number have to be of the form ****(2**^{p} − 1)(2^{p-1}) ?

The answer is yes, and here is my summary of a clearer version of the Euclid-Euler Theorem found here. Let the even perfect number = 2^{k} *a, where a is an odd number.

As seen earlier, if the number is perfect then

σ(2^{k} *a) = σ(2^{k}) σ(a) , which applies since a power of 2 and an odd number will never have a common factor greater than 1

As we demonstrated to get to equation 2(b), the sum of the factors of a power of 2 is equal to the next power of 2 minus 1, so

σ(2^{k} *a) = 2^{k+1} -1 * σ(a). ….equation (4)

Since 2^{k} *a is perfect, the sum of all of its factors will also equal twice itself, so:

σ(2^{k} *a) = 2*2^{k} *a = 2^{k+1} *a …equation (5)

Equating equations (4) and (5) we obtain,

(2^{k+1} -1)* σ(a) = 2^{k+1} *a …equation (6)

or rearranging σ(a) = 2^{k+1} *a /(2^{k+1} -1)

We can see from this expression that σ(a) is the product of 2^{k+1 }and another expression which we’ll simplify as* b*.

so σ(a) = 2^{k+1} *b , where b = a/ (2^{k+1} -1) …equation (7)

Rearranging the expression for b,

a = (2^{k+1} -1) * b and

σ(a) = σ( (2^{k+1} -1) * b) …..equation (8)

From equation (7)

σ(a) = 2^{k+1} *b , which we can rewrite as σ(a) =2^{k+1} *b -b + b . If we then factor the first two terms to get:

σ(a) = b(2^{k+1} -1) + b . ….equation (9)

Equating (8) and (9):

σ( (2^{k+1} -1) * b) = b(2^{k+1} -1) + b . The only way that the factors of (2^{k+1} -1) * b won’t exceed the discovered sum of b(2^{k+1} -1) + b is if *b* =1, so our original odd number *a* only has two factors, one of which is 1, revealing that** ***a* is prime. Moreover *a* is a Mersenne prime since it is of the form 2^{k+1} – 1

A contemporary of Mersenne, Pietro Cataldi discovered that perfect numbers end in either 6 or 8.

7. Finally we return to the question in the title. **Why do all even perfect numbers (no one yet found an odd-numbered one ) have to end in 6 or 8?**

A perfect number = ( 2^{p-1 })(2^{p} − 1). The first factor of that expression, 2^{p-1} , will always create an even power of 2, except for 2^{2-1} which will still satisfy our overall generalisation by creating 6. All even powers of 2 end either with the digit 6 or 4. The next power 2^{p} , which follows 2^{p-1 }, ends in either 2 or 8, respectively. But we subtract 1 from each of those numbers because we are multiplying ( 2^{p-1 }) by (2^{p} − 1). That translates into either multiplying a number ending in 6 by (1=2-1), which will end in 6 or a number ending in 4 by (7= 8-1), whose result will end in 8.

Here’s a table from Wikipedia of all known perfect numbers as of February 16, 2022. To date, thirty two of the discovered perfect numbers end in 6 and nineteen end in 8.

Rank |
*p for Mp* |
Perfect number |
Digits |
Year |
Discoverer |

1 |
2 |
6 |
1 |
4th century B.C.^{[5]} |
Euclid |

2 |
3 |
28 |
2 |
4th century B.C. |
Euclid |

3 |
5 |
496 |
3 |
4th century B.C. |
Euclid |

4 |
7 |
8128 |
4 |
4th century B.C. |
Euclid |

5 |
13 |
33,550,336 |
8 |
1456 |
First seen in a medieval manuscript, *Munich, Bayerische Staatsbibliothek, CLM 14908, fol. 33*^{[6]} |

6 |
17 |
8,589,869,056 |
10 |
1588 |
Cataldi^{[1]} |

7 |
19 |
137,438,691,328 |
12 |
1588 |
Cataldi^{[1]} |

8 |
31 |
2,305,843,008,139,952,128 |
19 |
1772 |
Euler |

9 |
61 |
265845599156…615953842176 |
37 |
1883 |
Pervushin |

10 |
89 |
191561942608…321548169216 |
54 |
1911 |
Powers |

11 |
107 |
131640364585…117783728128 |
65 |
1914 |
Powers |

12 |
127 |
144740111546…131199152128 |
77 |
1876 |
Lucas |

13 |
521 |
235627234572…160555646976 |
314 |
1952 |
Robinson |

14 |
607 |
141053783706…759537328128 |
366 |
1952 |
Robinson |

15 |
1,279 |
541625262843…764984291328 |
770 |
1952 |
Robinson |

16 |
2,203 |
108925835505…834453782528 |
1,327 |
1952 |
Robinson |

17 |
2,281 |
994970543370…675139915776 |
1,373 |
1952 |
Robinson |

18 |
3,217 |
335708321319…332628525056 |
1,937 |
1957 |
Riesel |

19 |
4,253 |
182017490401…437133377536 |
2,561 |
1961 |
Hurwitz |

20 |
4,423 |
407672717110…642912534528 |
2,663 |
1961 |
Hurwitz |

21 |
9,689 |
114347317530…558429577216 |
5,834 |
1963 |
Gillies |

22 |
9,941 |
598885496387…324073496576 |
5,985 |
1963 |
Gillies |

23 |
11,213 |
395961321281…702691086336 |
6,751 |
1963 |
Gillies |

24 |
19,937 |
931144559095…790271942656 |
12,003 |
1971 |
Tuckerman |

25 |
21,701 |
100656497054…255141605376 |
13,066 |
1978 |
Noll & Nickel |

26 |
23,209 |
811537765823…603941666816 |
13,973 |
1979 |
Noll |

27 |
44,497 |
365093519915…353031827456 |
26,790 |
1979 |
Nelson & Slowinski |

28 |
86,243 |
144145836177…957360406528 |
51,924 |
1982 |
Slowinski |

29 |
110,503 |
136204582133…233603862528 |
66,530 |
1988 |
Colquitt & Welsh |

30 |
132,049 |
131451295454…491774550016 |
79,502 |
1983 |
Slowinski |

31 |
216,091 |
278327459220…416840880128 |
130,100 |
1985 |
Slowinski |

32 |
756,839 |
151616570220…600565731328 |
455,663 |
1992 |
Slowinski & Gage |

33 |
859,433 |
838488226750…540416167936 |
517,430 |
1994 |
Slowinski & Gage |

34 |
1,257,787 |
849732889343…028118704128 |
757,263 |
1996 |
Slowinski & Gage |

35 |
1,398,269 |
331882354881…017723375616 |
841,842 |
1996 |
Armengaud, Woltman, et al. |

36 |
2,976,221 |
194276425328…724174462976 |
1,791,864 |
1997 |
Spence, Woltman, et al. |

37 |
3,021,377 |
811686848628…573022457856 |
1,819,050 |
1998 |
Clarkson, Woltman, Kurowski, et al. |

38 |
6,972,593 |
955176030521…475123572736 |
4,197,919 |
1999 |
Hajratwala, Woltman, Kurowski, et al. |

39 |
13,466,917 |
427764159021…460863021056 |
8,107,892 |
2001 |
Cameron, Woltman, Kurowski, et al. |

40 |
20,996,011 |
793508909365…578206896128 |
12,640,858 |
2003 |
Shafer, Woltman, Kurowski, et al. |

41 |
24,036,583 |
448233026179…460572950528 |
14,471,465 |
2004 |
Findley, Woltman, Kurowski, et al. |

42 |
25,964,951 |
746209841900…874791088128 |
15,632,458 |
2005 |
Nowak, Woltman, Kurowski, et al. |

43 |
30,402,457 |
497437765459…536164704256 |
18,304,103 |
2005 |
Cooper, Boone, Woltman, Kurowski, et al. |

44 |
32,582,657 |
775946855336…476577120256 |
19,616,714 |
2006 |
Cooper, Boone, Woltman, Kurowski, et al. |

45 |
37,156,667 |
204534225534…975074480128 |
22,370,543 |
2008 |
Elvenich, Woltman, Kurowski, et al. |

46 |
42,643,801 |
144285057960…837377253376 |
25,674,127 |
2009 |
Strindmo, Woltman, Kurowski, et al. |

47 |
43,112,609 |
500767156849…221145378816 |
25,956,377 |
2008 |
Smith, Woltman, Kurowski, et al. |

48 |
57,885,161 |
169296395301…626270130176 |
34,850,340 |
2013 |
Cooper, Woltman, Kurowski, et al. |

49 |
74,207,281 |
451129962706…557930315776 |
44,677,235 |
2016 |
Cooper, Woltman, Kurowski, Blosser, et al. |

50 |
77,232,915 |
109200…301056 |
46,498,850 |
2018 |
Cooper, Woltman, Kurowski, Blosser, et al. |

51 |
82,589,933* |
110847…207936 |
49,724,095 |
2018 |
Cooper, Woltman, Kurowski, Blosser, et al. |

The cover and an extract from Cataldi’s treatise on perfect numbers. The entire original is found at http://mathematica.sns.it/media/volumi/69/Trattato%20de’%20numeri%20perfetti_bw_.pdf

### Like this:

Like Loading...