Given the equation $w^2+x^2+y^2+z^2=wx+wy+wz+xy+xz+yz$, how does one systematically enumerate all non-negative integer solutions $\{w,x,y,z\}$?
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$\begingroup$ It's an easy equation. For me the interest is the idea to solve this equation: $x^2+y^2+z^2+q^2=xy+xz+xq+yz+yq+zq=r^2$ $\endgroup$– individOct 10, 2014 at 9:44
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$\begingroup$ For similar equations there a formula to look at it. math.stackexchange.com/questions/728994/… $\endgroup$– individOct 10, 2014 at 9:54
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2 Answers
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EEDDDDIIITTT: compare Diophantine quartic equation in four variables which discusses some of the same themes.
Very similar to Apollonian Gaskets. A reference for that is http://arxiv.org/abs/math/0009113
The main thing to realize is that, given any integer solution $(w,x,y,z),$ we get four neighboring solutions, $$ (x + y + z - w,x,y,z), $$ $$ (w, w + y + z - x,y,z), $$ $$ (w, x, w + x + z -y,z), $$ $$ (w, x,y, w + x + y -z). $$ There is a good deal more to say but my dinner is ready.
Later: ran on computer, it appears that the big difference from Apollonian is that we can take all entries positive, with the single (primitive) exception of the root $(0,1,1,1).$ More work to do.
Alright, I think already that this has a greater resemblance to http://en.wikipedia.org/wiki/Markov_number That is, if, for example, we always re-order so that $w \leq x \leq y \leq z,$ then the first three flips written above give new quadruples, while flipping the largest, $z,$ returns us to a previous quadruple (i.e. decreases $w+x+y+z$). That is, solutions make up a single tree, with a single root. However, in comparison to the Markov tree, there are plenty of quadruples with repeated entries.
Proof of all these apparent patterns is likely to be fairly long. I can tell you already, though, that a big part of it is showing that the only solution with $ \; \; 0 \leq w \leq x \leq y \leq z, \; \; \; \; \gcd(w,x,y,z) = 1,$ and $w + x + y \geq 2 z,$ is $(0,1,1,1).$ Usually but not always easier, one must show there are no solutions with some negative and some positive entries. False for the Apollonian case, appears true here. Big difference caused by just changing one multiplier from 2 to 1.
w x y z w + x + y - 2 z
0 1 1 1 0
1 1 1 3 -3
1 1 3 4 -3
1 3 4 7 -6
1 3 7 7 -3
1 4 7 9 -6
1 7 7 12 -9
1 7 9 13 -9
1 7 12 13 -6
1 9 13 16 -9
1 12 13 19 -12
1 12 19 19 -6
1 13 16 21 -12
1 13 19 21 -9
1 16 21 25 -12
1 19 19 27 -15
1 19 21 28 -15
1 19 27 28 -9
1 21 25 31 -15
1 21 28 31 -12
1 25 31 36 -15
1 27 28 37 -18
1 27 37 37 -9
1 28 31 39 -18
1 28 37 39 -12
1 31 36 43 -18
1 31 39 43 -15
1 36 43 49 -18
1 37 37 48 -21
1 37 39 49 -21
1 37 48 49 -12
1 39 43 52 -21
1 39 49 52 -15
1 43 49 57 -21
1 43 52 57 -18
1 48 49 61 -24
1 48 61 61 -12
1 49 52 63 -24
1 49 57 64 -21
1 49 61 63 -15
1 52 57 67 -24
1 52 63 67 -18
3 4 7 13 -12
3 4 13 13 -6
3 7 7 16 -15
3 7 13 19 -15
3 7 16 19 -12
3 13 13 25 -21
3 13 19 28 -21
3 13 25 28 -15
3 16 19 31 -24
3 16 31 31 -12
3 19 28 37 -24
3 19 31 37 -21
3 25 28 43 -30
3 25 43 43 -15
3 28 37 49 -30
3 28 43 49 -24
3 31 31 49 -33
3 31 37 52 -33
3 31 49 52 -21
3 37 49 61 -33
3 37 52 61 -30
3 43 43 64 -39
3 43 49 67 -39
3 43 64 67 -24
4 7 9 19 -18
4 7 13 21 -18
4 7 19 21 -12
4 9 19 25 -18
4 13 13 27 -24
4 13 21 31 -24
4 13 27 31 -18
4 19 21 37 -30
4 19 25 39 -30
4 19 37 39 -18
4 21 31 43 -30
4 21 37 43 -24
4 25 39 49 -30
4 27 31 49 -36
4 27 49 49 -18
4 31 43 57 -36
4 31 49 57 -30
4 37 39 61 -42
4 37 43 63 -42
4 37 61 63 -24
4 39 49 67 -42
4 39 61 67 -30
7 7 12 25 -24
7 7 16 27 -24
7 7 25 27 -15
7 9 13 28 -27
7 9 19 31 -27
7 9 28 31 -18
7 12 13 31 -30
7 12 25 37 -30
7 12 31 37 -24
7 13 19 36 -33
7 13 21 37 -33
7 13 28 39 -30
7 13 31 39 -27
7 13 36 37 -18
7 16 19 39 -36
7 16 27 43 -36
7 16 39 43 -24
7 19 21 43 -39
7 19 31 48 -39
7 19 36 49 -36
7 19 39 49 -33
7 19 43 48 -27
7 21 37 52 -39
7 21 43 52 -33
7 25 27 52 -45
7 25 37 57 -45
7 25 52 57 -30
7 27 43 61 -45
7 27 52 61 -36
7 28 31 57 -48
7 28 39 61 -48
7 28 57 61 -30
7 31 37 63 -51
7 31 39 64 -51
7 31 48 67 -48
7 31 57 67 -39
7 31 63 64 -27
7 36 37 67 -54
9 13 16 37 -36
9 13 28 43 -36
9 13 37 43 -27
9 16 37 49 -36
9 19 25 49 -45
9 19 31 52 -45
9 19 49 52 -27
9 25 49 64 -45
9 28 31 61 -54
9 28 43 67 -54
9 28 61 67 -36
12 13 19 43 -42
12 13 31 49 -42
12 13 43 49 -30
12 19 19 49 -48
12 19 43 61 -48
12 19 49 61 -42
12 25 37 67 -60
12 25 67 67 -30
13 13 25 48 -45
13 13 27 49 -45
13 13 48 49 -24
13 16 21 49 -48
13 16 37 57 -48
13 16 49 57 -36
13 19 21 52 -51
13 19 28 57 -54
13 19 36 61 -54
13 19 43 63 -51
13 19 52 63 -42
13 19 57 61 -33
13 21 31 61 -57
13 21 37 64 -57
13 21 49 67 -51
13 21 52 67 -48
13 21 61 64 -33
13 25 28 63 -60
13 27 31 67 -63
16 19 31 63 -60
16 19 39 67 -60
16 19 63 67 -36
16 21 25 61 -60
19 19 27 64 -63
19 21 28 67 -66
w x y z w + x + y - 2 z
jagy@phobeusjunior:
The fastest growth is along the branch where a quadruple is four consecutive values of the sequence $$ 1,3,4,7,13,21,37,64,109,189,325,559,964,1659, 2857,4921, \ldots $$ which satisfy $$ a_{n+4} = a_{n+3} + a_{n+2} + a_{n+1} - a_n. $$ The characteristic polynomial for this is $$ x^4 - x^3 - x^2 - x + 1 = \left( x^2 - \left(\frac{\sqrt {13}+1}{2} \right) x + 1 \right) \left( x^2 + \left(\frac{\sqrt {13}-1}{2} \right) x + 1 \right) $$ The quadratic factor on the left gives two real positive roots, the largest is $1.7220838...,$ the other smaller than 1. The quadratic factor on the right gives complex conjugate roots of magnitude exactly $1,$ so their contribution to $a_n$ does not completely die out but does stay bounded.
Alright, it all worked out. With $w \leq x \leq y \leq z$ a solution, we cannot have $w \leq x < 0 < y \leq z$ because $$ (w^2 - w x + x^2) + (y^2 - yz+z^2) = (w + x)(y + z). $$ This is a minor variant of an inequality on page 14 of Graham et al., link to arXiv above. Next, we cannot have $w = x + y + z = 0$ for a nontrivial solution, because $w = -(x+y+z)$ implies $3 (x^2 + y^2 + z^2 + yz+zx+xy ) = 0,$ but this is a positive definite form and thus $x=y=z=0$ and $w=0.$ Taking $w+x+y+z > 0$ as a hypothesis, the 1907 technique of Hurwitz on finding a Grundlösung ($w + x + y \geq 2 z$) with inequalities first give $w \geq 0,$ then $z=y=x$ and $w=0.$ In order to have the GCD turn into one we have $z=y=x=1.$
So, the whole thing really is a tree, single root, any solution can be reduced to $(0,1,1,1)$ in a finite number of steps(of the four types above), along with re-ordering at need.
Saturday: slight change, the thing is not quite a tree, although it remains a partially ordered set in which any (ordered) quadruple can be reduced to $(0,1,1,1)$ in a finite number of steps, each time strictly reducing $w+x+y+z.$ The surprise for me was that $(1,7,12,13)$ can reduced to either $(1,7,12,7),$ ordered $(1,7,7,12),$ or $(1,7,9,13),$ already ordered. Below are the first few layers of the poset, building off $(1,3,4,7):$
The first three dozen solution quadruples that reduce in two distinct ways, ordered by $w+x+y+z,$ are
33 1 7 12 13 -6
45 3 7 16 19 -12
51 4 7 19 21 -12
54 1 13 19 21 -9
66 7 7 25 27 -15
69 3 13 25 28 -15
75 1 19 27 28 -9
75 4 13 27 31 -18
75 7 9 28 31 -18
81 1 21 28 31 -12
87 7 12 31 37 -24
90 3 19 31 37 -21
90 7 13 31 39 -27
93 7 13 36 37 -18
99 4 19 37 39 -18
102 9 13 37 43 -27
105 1 28 37 39 -12
105 4 21 37 43 -24
105 7 16 39 43 -24
114 1 31 39 43 -15
114 7 19 39 49 -33
117 12 13 43 49 -30
117 7 19 43 48 -27
123 13 13 48 49 -24
123 3 28 43 49 -24
123 7 21 43 52 -33
129 9 19 49 52 -27
135 13 16 49 57 -36
135 1 37 48 49 -12
135 3 31 49 52 -21
141 12 19 49 61 -42
141 1 39 49 52 -15
141 4 31 49 57 -30
141 7 25 52 57 -30
147 13 19 52 63 -42
147 7 27 52 61 -36
In comparison, the original Apollonian Gasket really does give trees, any AG quadruple can be reduced only one way. Of course, there are infinitely many trees involved.
answered Oct 10, 2014 at 2:58
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Equation:
$$x^2+y^2+z^2+q^2=xy+xz+xq+yz+yq+zq$$
Solutions have the form:
$$x=3b^2(p-s)^2+a^2s^2$$
$$y=3b^2p^2+(a+3b)^2s^2$$
$$z=12b^2(p^2-ps+s^2)$$
$$q=(3bp+as)^2+3b^2s^2$$
Solutions have the form:
$$x=21b^2p^2+6b(a-b)ps+(a^2+6ab+21b^2)s^2$$
$$y=21b^2p^2+6b(a-3b)ps+(a-3b)^2s^2$$
$$z=3b^2p^2+6b(a+b)ps+3(a+b)^2s^2$$
$$q=9b^2p^2-6b(a+3b)ps+(a^2+6ab+21b^2)s^2$$
$a,b,p,s$ - integers asked us any sign.
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$\begingroup$ I am always impressed by your ability to solve those complex diophantine equations.Honestly, once I saw the question, I knew you would jot down the answer. Besides the obvious that you are a bright guy, I would like to know your methods. Is there any manual that you can recommend to me? $\endgroup$– user97615Oct 17, 2014 at 2:11
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$\begingroup$ @NumThcurious the methods I use are prohibited. No need to discuss it. You may have this problem. $\endgroup$– individOct 17, 2014 at 5:20
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$\begingroup$ prohibited methods? $\endgroup$– user97615Oct 17, 2014 at 18:07
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$\begingroup$ @NumThcurious mathisfunforum.com/viewtopic.php?id=20642 $\endgroup$– individOct 18, 2014 at 5:18
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$\begingroup$ I've visited your blogs several times before. You always provide answers to problems but never show your methods just like you did above. $\endgroup$– user97615Oct 18, 2014 at 10:35