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A Proof that Shoots its own Foot

Recently I began to get interested in odd perfect numbers. A perfect number is one that is equal to the sum of the divisors less than it. For example, 28 is divisible by 1, 2, 3, 4, 7, and 14 and these numbers sum to 28, so 28 is a perfect number. Euler proved that a number that is of the form 2ⁿ*(2ⁿ⁺¹-1) is perfect, and that all even perfect numbers are produced from this formula for some n. It is not known if an odd number can be perfect; it would have to satisfy so many conditions that it would make a lawyer's head swim. But no one has proved that there isn't one.

So it surprised me when I found one on the Internet! The number is 198,585,576,189. This number's factorization is 3²*7²*11²*13²*22,021. It can easily be shown that an odd perfect number has to be an odd prime times a square number. This number certainly has this form, provided of course that 22,021 is prime. It is not readily obvious that 22,021 is prime; it looks like it may be. The interesting thing about it is that if you assume that 22,021 is prime, then you can show that 198,585,576,189 is indeed a perfect number. But when doing this, you do something else that undermines this proof altogether. The proof that 198,585,576,189 is perfect also implies that 22,021 is not prime! In fact, it shows that 22,021 = 19²*61.

So here we have a case of proving the statement A -> B. But in so doing you also prove A -> ~A; i.e., that A implies not A, the statement that A is false. This destroys the proof altogether, because although it may be true that A -> B, that does not tell us anything, for if A -> ~A, then ~A must be true, and so A is false, so that A -> B is true but only because a false statement implies anything. It is the proof that shoots its own foot.

By the way, the page which shows the number 198,585,576,189 and shows that if 22,021 is assumed to be prime, then the number is perfect, is http://www.primepuzzles.net/puzzles/puzz_111.htm.