"Christian Boyer" <cboyer(a)club-internet.fr> wrote:
> About your conjecture 1:
> "Euler and Legendre demonstrated that x^3 + y^3 = k * z^3 is
> impossible with distinct integers, for k = 1, 2, 3, 4, 5."
> About your conjecture 2:
> "Legendre showed also that x^4 + y^4 = 2 * z^2 is impossible if x != y."
Thanks. What about higher powers? This is smelling a lot like
Fermat's Last Theorem, which is way above my pay grade.
Gareth McCaughan <gareth.mccaughan(a)pobox.com> wrote:
> Keith's conjecture 3 was: "No two positive integers have both a
> geometric mean that's an integer and a quadratic mean that's an
> integer." ....
Thanks for the proof that I guessed right (for values of "guess" that
consist of trying all pairs of distinct positive integers less than
ten thousand, about 50 million pairs).
I've since tested all power means from -5 through +5. It looks like
there are no integer means for any power whose absolute value is more
than 2. (Though admittedly I haven't tried powers of plus or minus
six or more, or pairs of integers where either integer was more than
9999.)
So it looks like there are just five kinds of power means that can
give integers, the -2, the harmonic (-1), geometric (0), arithmetic
(1), and quadratic (2). Is there a name for the -2 power mean? Until
someone can give me something better, I'll call it the citardauq mean.
Of those five possible means, there are 32 possible combinations
of means that are and aren't integers, for each pair of positive
integers. But apparently only 12 combinations happen. Here are
how many I get for each combination
24981771 00000 none
24954440 00010 arithmetic only
8638 00011 arithmetic and quadratic
8638 00100 geometric only
14118 00110 geometric and arithmetic
4322 01000 harmonic only
11565 01010 harmonic and arithmetic
338 01011 harmonic, arithmetic, and quadratic
269 01100 harmonic and geometric
524 01110 harmonic, geometric, and arithmetic
325 10011 citardauq, arithmetic, and quadratic
53 11011 citardauq, harmonic, arithmetic, and quadratic
The first example of the last has the following four means: 28, 35,
80, and 100. I'll leave it as an exercise for the reader to figure
out what the integers these are the means of are.
I had already noticed that the geometric and quadratic means are never
both integers. That was my conjecture 3. I now notice that the same
is true of the citardauq and geometric means; they never appear together.
Only two other rules are necessary. No quadratic means without
arithmetic means. That's easy to explain. Arithmetic means are
integers if and only if the numbers they're the means of have the
same parity (both odd or both even). Quadratic means also require
the parity be the same, since squares share the parity of what
they're the square of.
The final rule is that there's never a citardauq mean unless there's a
quadratic mean.
The quadratic mean, q, is defined as: (a^2 + b^2) / 2 = q^2
The citardauq mean, c, is defined as: (1/a^2 + 1/b^2)/2 = 1/c^2
Do some algebra, and get c^2 = (a^2 b^2)/(q^2), which shows that c
can't be an integer unless q is an integer. No, wait, no it doesn't;
q could be a rational number or the square root of a rational number.
Back to the drawing board. Sigh.
