## Perfect numbers and Pythagorean triples

For my first entry, let’s talk about a problem I posed in Mathematics Magazine several years ago.  (Plus it gives me a chance to mention Fermat, which I really should.)

Two of the more famous topics in elementary number theory are perfect numbers and Pythagorean triples.  A perfect number is one that is the sum of its proper divisors, excluding itself.  Examples include 6 ($6 = 1+2+3$) and 28 ($28 = 1 + 2 + 4 + 7 + 14$).  Euclid proved that if $2^p-1$ is prime, then $2^{p-1}(2^p-1)$ is perfect.  Euler proved that every even perfect number must be of this form.  Nobody has found an odd perfect number, yet no one has proved that they don’t exist, either.  The question of the existence of an odd perfect number is one of the more famous open problems in elementary number theory.

A Pythagorean triple consists of three positive numbers a, b, c such that $a^2 + b^2 =c^2$.  Examples include $(3,4,5)$ and $(5, 12, 13)$.  Sometimes a and b are called the legs of the triple and c is the hypotenuse.  Two of the more important results on Pythagorean triples are 1) all the primitive ones (those with no common factor) can be generated by the formula $a = 2mn, b = m^2-n^2, c = m^2+n^2$, where m and n are relatively prime positive integers of opposite parity; and 2) Every non-primitive Pythagorean triple is a multiple of a primitive one.

The proposed problem was to prove the following:

1. An even perfect number cannot be the hypotenuse of a Pythagorean triple.
2. If there is an odd perfect number, it must be the hypotenuse of a Pythagorean triple.

Proof of 1: Suppose c is even and perfect and is the hypotenuse of a Pythagorean triple.  Then there exists a primitive Pythagorean triple $(x,y,z)$ such that c is a multiple of z.  Since m and n in the generating formula for primitive Pythagorean triples must be positive and of different parity, z must be odd and greater than 1.  Thus $z = 2^p-1$, the only odd divisor of c greater than 1.  Since $p \geq 2$, $z \equiv 3 \bmod 4$.  However, one of m and n is even and the other odd.  The square of an odd number is congruent to $1 \bmod 4$, and the square of an even number is congruent to $0 \bmod 4$, so $z = m^2+n^2 \equiv 1 \bmod 4$.  Contradiction.

Proof of 2: Euler proved that any odd perfect number c must have a prime divisor p congruent to $1 \bmod 4$Fermat’s theorem on sums of two squares says that an odd prime number can be expressed as the sum of two squares if and only if it is congruent to $1 \bmod 4.$  Thus there exist integers m and n such that $m^2 + n^2 = p$.  Since p is an odd prime, m and n must have opposite parity.  Assume, without loss of generality, that $m > n.$  Let $x = 2mn, y = m^2-n^2$.  Thus $(x,y,p)$ is a Pythagorean triple.  Therefore, $(xc/p, yc/p,c)$ is also a Pythagorean triple, making c the hypotenuse of a Pythagorean triple.

At first glance this result seems like it gives some weight to the “no odd perfect numbers” side of the open question, rather than the “we just haven’t found one yet” side.  However, since it depends so heavily on the standard result of Euler’s on odd perfect numbers I don’t think that it really advances our knowledge any on this question.  It’s more of a “huh, isn’t that interesting” result – which makes it appropriate for the problem section of a journal rather than the research article section.

References:

1. Michael Z. Spivey, Problem 1737, Mathematics Magazine, 79 (1): 67, February 2006.
2. Robert L. Doucette, Solution to Problem 1737, Mathematics Magazine, 80 (1): 79-80, February 2007.
3. Pythagorean Triples and Perfect Numbers,” at Cut-the-Knot.