Hi, Pete. There are a few observations related to this, not widely known although basic, and that includes your colleague. First, Conway gives a quick proof on page 142 of The Sensual Quadratic Form, including over the rationals.
Next, also Conway, the form (five variables) that he and Schneeberger found that represents all the numbers from 1 to 289, fails to represent 290, then represents 291 and on forever, he initially called Methusaleh. It is just a binary added to a ternary that represents the numbers from 1 to 28 consecutive, discriminant 29. However, for ternaries that is not the record. The form he called Little Methusaleh, discriminant 31, represents 1 to 30 consecutive. The theorem is in this material, as the conditions for a positive ternary to represent, say, 1,2,3,5, places strong restrictions on a partly reduced form. Kap wrote this sort of argument up several times, including a repeat in the unpublished 1996 Classification. It is quite easy. OK, Little Methusaleh and your result over the integers are proved on page 81 of The Sensual Quadratic Form
Finally, a positive form is anisotropic at the "prime" infinity. In Cassels Rational Quadratic Forms he shows global relations on the Hilbert Norm Residue symbol that show that any ternary is anisotropic at an even number of primes. So a positive ternary is anisotropic at an odd number of finite primes. Taken with the observation above that at least one number below 31 is missed, and a positive ternary fails to integrally represent an infinite number of positive integers.
I will look up some of my tables and fill things in. Note that some of this is discussed in an early article by William Duke, 1997 Notices, but he mistyped the form with discriminant 29.
Let's see, Conway and Schneeberger probably had an acceptable proof of the 15 Theorem scattered about, but it never got put together. Bhargava was looking for diversions from his own dissertation, Conway mentioned this in passing. Bhargava showed the fundamental result that one of these forms must have a regular ternary as a sub-form, thus the project became a careful inspection of my paper with Kap on all possible regular ternaries. Also, correspondence between Kap and Bhargava first revealed some important errors in Magma relating to calculating the spinor genus, and hilarity ensued.
EDIT: thinking about the history question, it is quite possible that this result was never written down as a separate proposition, by Gauss, Legendre, etc. The reason I suggest this is the great weight placed on positive ternary forms missing certain "progressions," in the language of Jones, Dickson, other early books. So, in Jones, chapter 8, we read "Thus there will be a finite number of arithmetical progressions of this type" of numbers not represented by any form in the genus under consideration. Not much motivation for proving that a form misses at least one number if you are going to quickly show that it misses an entire arithmetic progression.
EDIT TOOO: note that Conway replaces the prime usually called $\infty$ by the prime $-1.$
No definite ternary form is universal
However, a simple argument shows that
any definite ternary form must fail to
represent infinitely many integers,
even over the rationals. For if a
ternary form $f$ of determinant $d$
represents anything in the $p$-adic
squareclass of $-d$ over $\mathbf
> Q_p,$ then it must be $p$-adically
equivalent to $[ -d,a,b]$ where the
"quotient form" $[a,b]$ has
determinant $-1,$ and so $p$-adically,
$f$ must be the isotropic form $[
> -d,1,-1].$
But a positive definite form fails to
represent $-1,$ and so it is not
$p$-adically isotropic for $p=-1.$ By
the global relation, there must be
another $p$ for which it is not
$p$-adically isotropic, and so it
also fails to represent all numbers in
the $p$-adic square-class of $-d$ for
this $p$ too!
The Three Squares Theorem illustrates
this nicely--the form $[ 1,1,1]$ fails
to represent $-1$ both $-1$-adically
and $2$-adically. In the Third
Lecture, we showed that The Little
Methusaleh Form $$ x^2 + 2 y^2 + y z
> + 4 z^2 $$ fails to represent 31. We now see that since it fails to
represent the $-1$-adic class of its
determinant $-31/4$ (i.e., the
negative numbers), it must also fail
to represent the infinitely many
positive integers in the $31$-adic
squareclass of $-31/4.$
Disclaimer: This is not a complete solution but some partial results that may be helpful for a complete solution. Also note that this is my first contribution to Math Overflow so I hope I follow all guidelines and I would appreciate tolerance and feedback if I did something wrong.
I couldn't find any further progress on this problem on the internet (other than this thread). I believe I can reduce the search space to about $\approx 3.5*10^8$ cases.
Any powerful number can be written in the form $mz^2$ with $m|z$, so we are trying to solve $x^4 + y^4 = mz^2$. Since $m|z, x^4 + y^4 > m^3$, we have to understand what happens for all $m < N^{1/3} ~ 1.5 * 10^{12}$.
Now $m$ must obey the following conditions (this is how I reduce the search space):
- $m$ must be squarefree.
- $m$ must be odd. Since $gcd(x,y) = 1, x^4 + y^4$ is either $1$ or $2 \mod 16$, but a powerful number is clearly not $2 \mod 16$.
- every prime factor $p$ of $m$ must be $1 \mod 8$, since if $p$ divides $x^4 + y^4 $but not $x$ or $y$, then $(x/y)^4 \equiv -1 \mod p$, so $p \equiv 1 \mod 8$.
- $2m$ must be a congruent number. Indeed, suppose $(x,y,z)$ is a solution to $x^4 + y^4 = mz^2$. Let $u = -4x^2y^2/z^2$, and let $v = 4xy(x^4-y^4)/z^3$. Then $v^2 = u^3-4m^2u$, so the congruent number curve $E(2m)$ is a principal homogenous space for the quartic $x^4 + y^4 = mz^2$.
Therfore we need to check curves with those odd $m$ such that each prime factor of $m$ is congruent to $1 \mod 8$, such that $2m$ is a congruent number which is $2 \mod 16$.
It is also known that there are no relatively prime solutions to $x^4 + y^4 = m^3$ (By Proposition 14.6.6 of Cohen's Number Theory: Volume II: Analytic and Modern Tools).
The next possibility, is $x^4 + y^4 = 17^2 m^3$. This means, we only need to check $m$ up to $(N/17^2)^{1/3} \approx 2.2 * 10^{11}$, i.e., $2m$ up to $\approx 4.4* 10^{11}$. By Table 3 of http://homepages.warwick.ac.uk/~masfaw/congruent.pdf (which counts the number of congruent numbers which are $2 \mod 16$ up to $10^{12}$), there are $59536672 + 62455317 + 66579936 + 73445274 + 81759844 + 12294626 + 2110645 = 358182314$ curves to check.
Best Answer
Here is a proof of the conjecture. I will refer several times to the book Cassels: Rational quadratic forms (Academic Press, 1978).
1. Let $p$ be a prime such that $p\nmid a$. Using the invertible linear change of variables over $\mathbb{Z}_p$ $$x'=ax+bz,\qquad y'=y+(c/a)z,\qquad z'=(1/a)z,$$ we have $$x'y'-(abc)z'^2=axy+byz+czx.$$ Therefore, the quadratic forms $axy+byz+czx$ and $xy-(abc)z^2$ are equivalent over $\mathbb{Z}_p$. By symmetry, we draw the same conclusion when $p\nmid b$ or $p\nmid c$ (note that $p$ cannot divide all of $a,b,c$).
2. For $p>2$, we see that $axy+byz+czx$ is equivalent to $x^2-y^2-(abc)z^2$ over $\mathbb{Z}_p$. Following the notation and proof of the first Corollary on p.214, we infer that $U_p\subset\theta(\Lambda_p)$. For $p=2$, we infer the same by the second Corollary of p.214. Now, combining the Corollary on p.213 with Theorem 1.4 on p.202, we conclude that the genus of $axy+byz+czx$ contains precisely one proper equivalence class.
3. By the conclusions of the previous two points, the quadratic forms $axy+byz+czx$ and $xy-(abc)z^2$ are properly equivalent. As $xy-(abc)z^2$ clearly represents all integers, the same is true of $axy+byz+czx$.
Remark. The crux of the proof are the Corollary on p.213 and Theorem 1.4 on p.202. The first statement relies on the Hasse principle (cf. Lemma 3.4 on p.209 and its proof). The second statement is a straightforward application of strong approximation for the spin group.