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 Kap wrote this sort of argument up several times, including a repeat in the unpublished 1996 Classification.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 $-31/4.$
Borovoi does not claim what you seem to think. Siegel's formalism discusses, for positive forms, the number of representations of an integer by an entire genus of positive ternary forms, each form weighted according to the number of its integral automorphs. For indefinite forms, the automorph groups are infinite and one must discuss the number of essentially inequivalent representations of an integer by a form, that is representations that cannot be taken to each other by automorphs.
That being said, most indefinite ternary forms are in a genus with only one integer equivalence class of forms. In such cases, Siegel's answers for a genus agree with those for the single form.
Borovoi simply repeats one of the more famous examples of a genus of more than one class. In the first edition (1988) of SPLAG by Conway and Sloane, this is on pages 404-405. The two classes are represented by
$$ x^2 - 2 y^2 + 64 z^2 $$ and
$$ -9 x^2 + 2 x y + 7 y^2 + 2 z^2. $$
In at least three variables, with indefinite forms, spinor genera coincide with equivalence classes. That is, every indefinite ternary form is spinor regular. There is, as there is here, a finite set of squareclasses of "spinor exceptional integers" that may fail to be represented by a form, even though another form in the genus represents them. Borovoi does not mention it in this manner (he takes only my $ \; s=1$), but here we go:
Let us define a (positive) integer $s$ such that all prime factors $q$ of $s$ satisfy
$$ q \equiv \pm 1 \pmod 8, $$
with the consequence that also $ s \equiv \pm 1 \pmod 8. $
Then, in notation going back to Jones and Pall (1939), Borovoi's form does not integrally represent $s^2,$ as in
$$ -9 x^2 + 2 x y + 7 y^2 + 2 z^2 \neq s^2.$$
Borovoi gives all necessary steps. I should say already that every example I know of spinor exceptional integers can be proved, after the fact, by very simple factoring arguments.
Proof (Zagier): ASSUME
$$ -9 x^2 + 2 x y + 7 y^2 + 2 z^2 = s^2$$ in integers, where
$$ q | s \; \; \Rightarrow q \equiv \pm 1 \pmod 8$$ Rewrite as
$$ 2 z^2 - s^2 = 8(x^2 - y^2) + (x-y)^2. $$
The left hand side is odd, the right side is also odd, so $x-y$ is odd, and the right side is $1 \pmod 8.$ If $z$ were even, the left side would be $7 \pmod 8,$ so we find that $z$ is odd. As a result, $z^2 \equiv 1 \pmod 8$ and $2 z^2 \equiv 2 \pmod {16}.$ As $s \equiv \pm 1 \pmod 8,$ we get $s^2 \equiv 1 \pmod {16}.$ Put together, we get
$$ 2 z^2 - s^2 = 1 \pmod {16}. $$
But $8 (x^2-y^2) \equiv 8 \pmod {16}.$ Now we find
$$ (x-y)^2 \equiv 9 \pmod {16}.$$ Backing this up we get
$$ x-y \equiv \pm 3 \pmod 8.$$ Thus there is some prime $r \equiv \pm 3 \pmod 8$ such that
$ r | (x-y).$ But, from $ 2 z^2 - s^2 = 8(x^2 - y^2) + (x-y)^2,$ we see $$ (x-y) | (2 z^2 - s^2). $$ Put those together, we have a prime $r$ such that
$$ r | (2 z^2 - s^2) \; \; \mbox{with} \; \; r \equiv \pm 3 \pmod 8. $$ By a standard application of quadratic reciprocity, we find that
$$ r | s $$
which is a contradiction of the assumption. Just for completeness, if $r$ does not divide $s,$ then $s$ has a multiplicative inverse $\pmod r.$ So $2 z^2 - s^2 \equiv 0 \pmod r$ becomes $2 z^2 \equiv s^2 \pmod r,$ then $4 z^2 \equiv 2 s^2 \pmod r,$ finally
$\left( \frac{2z}{s} \right)^2 \equiv 2 \pmod r.$ However, $2$ is not a quadratic residue $\pmod r.$
So, in fact,
$$ -9 x^2 + 2 x y + 7 y^2 + 2 z^2 \neq s^2.$$
As to the forms being in the same genus, write
$$ f(x,y,z) = x^2 - 2 y^2 + 64 z^2 $$
and
$$ g(x,y,z) = -9 x^2 + 2 x y + 7 y^2 + 2 z^2. $$
Then
$$ g(3 v + 15 w, u + v + 7 w, u - 5 v - 32 w) = 9 f(u,v,w). $$
Also
$$ f(u + 7 v + 2 w, 13 u -5 v + 5 w, 2 u - v + w) = 9 g(u,v,w). $$
Other aspects: with prime $r \equiv \pm 3 \pmod 8,$ we do have $f(r,0,0) = r^2.$ Another factoring argument shows that $f$ does not primitively represent $r^2.$ That is, if $f(x,y,z) = r^2,$ then $\gcd (x,y,z) = r.$ Finally, I am less sure about this, but I believe that $g$ represents all such $r^2,$ as in $g(x,y,z) = r^2,$ in this case primitively as $g$ does not represent 1. Examples include
$$ g(0,1,1) = 9, \; g(0,1,3) = 25, \; g(1,2,7) = 121,\; g(0,1,9) = 169, \; g(0,7,3) = 361, $$
$$ g(3,10,9) = 841, \; g(3,-14,9) = 1369, \; g(1,-16,7) = 1849,\; g(2,19,11) = 2809.$$
The shortest discussion on this is on page 352 of Rainer Schulze-Pillot, Exceptional Integers for Genera of Integral Ternary Positive Definite Quadratic Forms, Duke mathematical Journal, volume 102 (2000), pages 351-357. However, as in the title, he is talking about definite forms. There is also his survey, including Siegel for indefinite forms, Representation by integral quadratic forms-a survey, in Contemporary Mathematics volume 344 (2004) pages 303-321, the book is titled Algebraic and Arithmetic Theory of Quadratic Forms edited by Baeza, Hsia, Jacob, and Prestel.
Best Answer
Thursday: here is an example I proved in full detail, that illustrates the use of the mappings in one direction, along with the possible intricacy of the difference between finding all rational null vectors and successfully finding all primitive integer null vectors:
All solutions to $$ 2(x^2 + y^2 + z^2) - 113 (yz+zx+xy) = 0 $$ with $$ \gcd(x,y,z) = 1 $$ can be written as one of four recipes, with the understanding that we sort by absolute value and possibly multiply through by $-1$ so as to demand $x \geq |y| \geq |z|,$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 37 u^2 + 51 uv + 8 v^2 \\ 8 u^2 -35 uv -6 v^2 \\ -6 u^2 + 23 uv + 37 v^2 \end{array} \right) $$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 32 u^2 + 61 uv + 18 v^2 \\ 18 u^2 -25 uv -11 v^2 \\ -11 u^2 + 3 uv + 32 v^2 \end{array} \right) $$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 38 u^2 + 45 uv + 4 v^2 \\ 4 u^2 -37 uv -3 v^2 \\ -3 u^2 + 31 uv + 38 v^2 \end{array} \right) $$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 29 u^2 + 63 uv + 22 v^2 \\ 22 u^2 -19 uv -12 v^2 \\ -12 u^2 -5 uv + 29 v^2 \end{array} \right) $$
In all four cases we simply discard occurrences when the resulting $x,y,z$ have a common factor.
The four are all of the form $X = R U,$ where $$ X = \left( \begin{array}{r} x \\ y \\ z \end{array} \right) $$ and $$ U = \left( \begin{array}{r} u^2 \\ uv \\ v^2 \end{array} \right). $$ Clearly we take $\gcd(u,v) = 1.$ We can also take $u,v \geq 0.$ This is an artifact of the extreme symmetry of the ternary form and the extremely special form of the four matrices $R$ that I chose.
hi
well, then. I put in some blank lines..