When does $axy+byz+czx$ represent all integers?

Question

For which $a,b,c$ does $axy+byz+czx$ represent all integers?

In a recent answer, I conjectured that this holds whenever $\gcd(a,b,c)=1$, and I hope someone will know. I also conjectured that $axy+byz+czx+dx+ey+fz$ represents all integers when $\gcd(a,b,c,d,e,f)=1$ and each variable appears non-trivially, though I'm less optomistic about finding prior results on that. Here are some results:

• If $\gcd(a,b)=1$ then $axy+byz+czx$ represents all integers. [Proof: Find $r,s$ with $ar+bs=1$, then take $x = r$, $y = n - crs$, $z = s$.]

• $6xy+10yz+15zx$, the first case not covered above, represents all integers up to 1000. Similarly $77xy+91yz+143zx$ represents all integers up to 100. [by exhaustive search]

• If $\gcd(a,b,c)=1$ then $axy+byz+czx$ represents all integers mod $p^r$. [proved in the above link]

The literature on this is hard to search because these are not positive definite forms, and many apparently relevant papers only consider the positive definite case. For old results, the most relevant parts of Dickson's History of the Theory of Numbers (v. 2, p. 434; v. 3, p. 224) mention only the case of $xy+xz+yz=N$. Does anyone here know a general result or reference?

Answer 1

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.

Answer 2

Just so you know, one of Dickson's students (A. Oppenheim) finished classifying (indefinite) universal ternaries; the final family is $xy - M z^2.$ Page 161 in Modern Elementary Theory of Numbers. Your conjecture is that $xy-(abc) z^2$ is $SL_3 \mathbb Z$ equivalent to $ayz + b zx + c xy.$

For example, taking $$ u = 192x + 50 y + 45 z,$$

$$ v = 75 x + 18 y + 20 z, $$

$$ w = 4x + y + z, $$

$$ uv - 900 w^2 = 10 yz + 15 zx + 6xy $$

This is an equivalence (determinant $\pm 1$), one may invert the change of variables so that $10 yz + 15 zx + 6xy $ really is universal

Answer 3

I have figured out some things; it is much quicker, as far as computing, to find a way for the Hessian matrix of the ternary quadratic form, is to have it represent the (two by two) Hessian of the form $xy;$ this form, or its quadratic space, is often called The Hyperbolic Plane; see page 15 in Cassels.

Once this is done, there is just the business of adding an appropriate third row to "rows" to get a nice result. The final quadratic form is (y - 1250*z)*x + (-797*z*y - 5751*z^2) $$ xy -797yz - 1250 zx - 5751 z^2, $$ which is universal because we can take $z = 0, y = 1,$ and $x$ equal to the target number. Oh, the beginning form was your $$ 77yz + 91 zx + 143xy $$

$$ \left( \begin{array}{ccc} 830 &-3486 &-2145 \\ 616& -2587 & -1592 \\ -3& -5& 12 \\ \end{array} \right) \left( \begin{array}{ccc} 0 &143 &91 \\ 143& 0&77 \\ 91&77 &0 \\ \end{array} \right) \left( \begin{array}{ccc} 830 &616 &-3 \\ -3486& -2587 &-5 \\ -2145&-1592 &12 \\ \end{array} \right) = \left( \begin{array}{ccc} 0&1 &-1250 \\ 1&0 & -797 \\ -1250&-797 &-11502 \\ \end{array} \right) $$ Note: it turns out to be quite easy, with explicit matrices, to take the form with the visible hyperbolic plane to the form $xy - (abc) z^2,$ that GH has already proved equivalent to the original $ayz+bzx+cxy.$

$$ \left( \begin{array}{ccc} 830 &-3486 &-2145 \\ 616& -2587 & -1592 \\ 1431507& -6012097& -3699553 \\ \end{array} \right) \left( \begin{array}{ccc} 0 &143 &91 \\ 143& 0&77 \\ 91&77 &0 \\ \end{array} \right) \left( \begin{array}{ccc} 830 &616 & 1431507 \\ -3486& -2587 &-6012097 \\ -2145&-1592 & -3699553 \\ \end{array} \right) = \left( \begin{array}{ccc} 0&1 &0 \\ 1&0 & 0 \\ 0& 0 &-2004002 \\ \end{array} \right) $$

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parisize = 4000000, primelimit = 500000
? h = [ 0,143,91; 143,0,77; 91,77,0]
%1 = 
[  0 143 91]

[143   0 77]

[ 91  77  0]

? rows = [ 830, -3486, -2145; 616, -2587, -1592 ]
%2 = 
[830 -3486 -2145]

[616 -2587 -1592]

? columns = mattranspose(rows)
%3 = 
[  830   616]

[-3486 -2587]

[-2145 -1592]

? rows * h * columns
%4 = 
[0 1]

[1 0]

? 
? 
? 
? rows = [ 830, -3486, -2145; 616, -2587, -1592; -3,-5,12 ]
%5 = 
[830 -3486 -2145]

[616 -2587 -1592]

[ -3    -5    12]

? matdet(rows)
%6 = 1
? columns = mattranspose(rows)
%7 = 
[  830   616 -3]

[-3486 -2587 -5]

[-2145 -1592 12]

? rows * h * columns
%8 = 
[    0    1  -1250]

[    1    0   -797]

[-1250 -797 -11502]

? x
%9 = x
? y
%10 = y
? z
%11 = z
? g = rows * h * columns
%12 = 
[    0    1  -1250]

[    1    0   -797]

[-1250 -797 -11502]

? vec = [ x,y,z]
%13 = [x, y, z]
? vect = mattranspose(vec)
%14 = [x, y, z]~
? vec * g * vect / 2 
%15 = (y - 1250*z)*x + (-797*z*y - 5751*z^2)
? 

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