Using your Head is Permitted

May 2009 riddle

This month's riddle deals with a well-known result in number theory. As always, we ask for original solutions only, not ones based on prior familiarity with the proof or on Internet searching.

In number theory, a perfect number is a number equal to the sum of all its proper divisors. An example of a perfect number is 28, because 28 equals 1+2+4+7+14.

28, specifically, is a number that can be represented in a special form, namely it can be represented as (p2+p)/2 where p is a Mersenne prime (a prime number that can be represented as 2n-1). For 28, p=7. We will call this representation "Special Form A". It is not difficult to see that any number that can be represented in special form A is a perfect number. However, it is currently an open problem whether there are any numbers not representable in this form that are also perfect numbers.

Consider, now, another special form, which we will call "Special Form B". It is (2m+1)2q, where m is a natural and q is a prime congruent to 1 (mod 4).

This month's riddle: prove that all perfect numbers must be either of special form A or of special form B.

List of solvers:

Oded Margalit (3 May 14:58)
Dan Dima (5 May 00:02)
Albert Stadler (7 May 21:20)
James Ge (14 May 17:56)
Itsik Horovitz (19 May 07:58)

Elegant solutions can be submitted to the puzzlemaster at Names of solvers will be posted on this page. Notify if you don't want your name to be mentioned.

The solution will be published at the end of the month.


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