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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?

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  • $\begingroup$ You mention "represents all integers up to 1000" but do you also include negative integers? Your title seems to indicate that all integers are to be represented including negative integers. $\endgroup$ – Somos Sep 17 at 20:11
  • $\begingroup$ Does $xy+yz+zx$ miss integer? $\endgroup$ – VS. Sep 17 at 20:32
  • $\begingroup$ @Somos, this includes negative integers too. The general results I quoted apply to both positive and negative numbers; and by GH’s answer below, the two specific polynomials also represent all integers. $\endgroup$ – Matt F. Sep 17 at 20:42
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    $\begingroup$ @VS. perhaps you are thinking of x,y,z positive and distinct: " A positive integer n is idoneal if and only if it cannot be written as ab + bc + ac for distinct positive integer a, b, and c" Excerpted from en.wikipedia.org/wiki/Idoneal_number $\endgroup$ – Will Jagy Sep 30 at 0:52
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    $\begingroup$ @VS. short proof at oeis.org/A000926/a000926.txt $\endgroup$ – Will Jagy Sep 30 at 0:59
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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.

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  • $\begingroup$ never noticed 1.5 $\endgroup$ – Will Jagy Sep 17 at 18:59
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    $\begingroup$ Thanks! I am getting the book from the library soon so I can make sense of this. $\endgroup$ – Matt F. Sep 17 at 21:06
  • $\begingroup$ @MattF. Thanks for the feedback. I should add that I find the conjecture very nice. $\endgroup$ – GH from MO Sep 17 at 21:18
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    $\begingroup$ GH, very, very nice. @MattF. I am fiddling, computer experiments, seeing if we can arrange your form equivalent to some $$ M z^2 + P yz + Q zx + xy, $$ which is universal because we can take $z=0$ and $y=1,$ finally $x$ set equal to the target number. Let's see, this discriminant comes out $PQ - M.$ I should write a separate program, my general purpose program for this is getting too slow for this problem. $\endgroup$ – Will Jagy Sep 20 at 18:57
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    $\begingroup$ @MattF. For $p=2$ you only need the Corollary on p.214, not its proof. The Corollary says that if $U_2\not\subset\Lambda(\Lambda_2)$, then $\phi$ is $2$-adically integral in the classical sense. The form $axy+byz+czx$ is not $2$-adically integral in the classical sense, because one of the coefficients $a,b,c$ is odd. Hence $U_2\subset\Lambda(\Lambda_2)$ holds for $axy+byz+czx$, as I claimed above. $\endgroup$ – GH from MO Sep 26 at 13:03
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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

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    $\begingroup$ I hoped you would see this, since I figured you were the person most likely to know — thanks! $\endgroup$ – Matt F. Sep 17 at 18:24
  • $\begingroup$ Where does this leave the answer to my original question? $\endgroup$ – Matt F. Sep 17 at 18:45
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    $\begingroup$ @MattF. it makes it a reasonable search for patterns in integral changes of variables between two specific forms. For example, coefficient triples $(1,1,p)$ and $(1,p,q)$ work, recognizable patterns. I need to get some groceries $\endgroup$ – Will Jagy Sep 17 at 18:54
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    $\begingroup$ @GHfromMO thanks for letting me know. I've been fiddling with examples. $\endgroup$ – Will Jagy Sep 17 at 21:09
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    $\begingroup$ I prove now the $\mathrm{SL}_3(\mathbb{Z})$ equivalence of $axy+byz+czx$ and $xy-(abc)z^2$ directly. See my updated post! $\endgroup$ – GH from MO Sep 18 at 17:42
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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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  • $\begingroup$ Another question you might like is: given $ap+bq+cr=1$, can we find $k$ relatively prime to $2abc$ such that for each $N$, the equation $axy+byz+czx=k^2N$ is solvable? Cassels proves a more general version (chapter 9, theorem 8.1) using more general theory, and this more general version goes into the results quoted by GH. I am looking for a simpler argument for this case, ideally with explicit polynomials for $k$ in terms of $a,b,c,p,q,r$, and for $x,y,z$ in terms of $a,b,c,p,q,r,k,N$. $\endgroup$ – Matt F. Sep 26 at 19:30
  • $\begingroup$ E.g. $77xy+91yz+143zx=907^2 n$ can always be solved by $x=21n+78$, $y=-26n+33$, $z = 66n-14$. $\endgroup$ – Matt F. Sep 27 at 22:09

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