How do you find out perfect numbers? - Answers

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, 13466917Note 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 to1050 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!6,28,496,8128,33550336,8589869056,137438691328,2305843008139952128,2658455991569831744654692615953842176,191561942608236107294793378084303638130997321548169216,13164036458569648337239753460458722910223472318386943117783728128,14474011154664524427946373126085988481573677491474835889066354349131199152128,23562723457267347065789548996709904988477547858392600710143027597506337283178622239730365539602600561360255566462503270175052892578043215543382498428777152427010394496918664028644534128033831439790236838624033171435922356643219703101720713163527487298747400647801939587165936401087419375649057918549492160555646976,141053783706712069063207958086063189881486743514715667838838675999954867742652380114104193329037690251561950568709829327164087724366370087116731268159313652487450652439805877296207297446723295166658228846926807786652870188920867879451478364569313922060370695064736073572378695176473055266826253284886383715072974324463835300053138429460296575143368065570759537328128,54162526284365847412654465374391316140856490539031695784603920818387206994158534859198999921056719921919057390080263646159280013827605439746262788903057303445505827028395139475207769044924431494861729435113126280837904930462740681717960465867348720992572190569465545299629919823431031092624244463547789635441481391719816441605586788092147886677321398756661624714551726964302217554281784254817319611951659855553573937788923405146222324506715979193757372820860878214322052227584537552897476256179395176624426314480313446935085203657584798247536021172880403783048602873621259313789994900336673941503747224966984028240806042108690077670395259231894666273615212775603535764707952250173858305171028603021234896647851363949928904973292145107505979911456221519899345764984291328You 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 knowsigma(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 sosigma(2k-1m) = sigma(2k-1)(sigma(m)) = (2k-1)(sigma(m)).Since n is perfect we also know thatsigma(n) = 2n = 2km.Together these two criteria give2km = (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 that2kM = 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.