# One Diophantine equation

I wonder now that the following Diophantine equation:

$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$

have only this formula describing his decision?

$a=-(k^2+2(p+s)k+p^2+ps+s^2)$

$b=2k^2+4(p+s)k+3p^2+3ps+2s^2$

$c=3k^2+4(p+s)k+2p^2+ps+2s^2$

$d=2k^2+4(p+s)k+2p^2+3ps+3s^2$

$k,p,s$ - what some integers.

By your question, I mean what that formula looks like this. Of course I know about the procedure of finding a solution, but I think that the formula would be better.

• en.wikipedia.org/wiki/… The solutions are a forest of countably many rooted trees. See articles in the AMS Bulletin by Fuchs and Kantorovich. – Will Jagy Mar 27 '14 at 8:25
• I understand that, but I was interested in something else. That is the formula itself. – individ Mar 27 '14 at 8:29
• no formula describes all solutions. Also, you don't seem to have asked any question. – Will Jagy Mar 27 '14 at 8:31
• As this is not the formula? What do you say? And this formula that I wrote? – individ Mar 27 '14 at 8:33
• I do not understand most of what you say; I suppose English is not your first language. It is certainly possible to describe some solutions of Apollonian circle packing with formulas. There will always be many other solutions that are not described by those formulas. – Will Jagy Mar 27 '14 at 9:06

I don't think your formula exhaust all solutions of the Diophantine equation

$$2(a^2+b^2+c^2+d^2) = (a+b+c+d)^2\tag{*1}$$

Your equation can be rewritten as

$$(a+b+c+d)^2 = (-a+b-c+d)^2 + (-a+b+c-d)^2 + (-a-b+c+d)^2$$

This is the equation for a Pythagorean quadruple. The set of all Pythagorean quadruples can be parametrized by 5 integers $\alpha,\beta,\gamma,\delta$ and $\lambda$:

$$\begin{cases} \hphantom{-}a + b + c + d &= \lambda (\alpha^2 + \beta^2 + \gamma^2 + \delta^2)\\ -a + b - c + d &= \lambda(\alpha^2+\beta^2 - \gamma^2 - \delta^2)\\ -a + b + c - d &= 2\lambda(\alpha\delta + \beta\gamma)\\ -a - b + c + d &= 2\lambda(\beta\delta - \alpha\gamma) \end{cases}$$ Using this, we see all solutions of $(*1)$ must have the form $$\begin{cases} a &= \frac{\lambda}{2} \left(\gamma^2 + \delta^2 + (\alpha-\beta)\gamma - (\alpha+\beta)\delta\right)\\ b &= \frac{\lambda}{2} \left(\alpha^2 + \beta^2 + (\alpha+\beta)\gamma + (\alpha-\beta)\delta\right)\\ c &= \frac{\lambda}{2} \left(\gamma^2 +\delta^2 - (\alpha-\beta)\gamma + (\alpha+\beta)\delta\right)\\ d &= \frac{\lambda}{2} \left(\alpha^2+\beta^2 - (\alpha+\beta)\gamma - (\alpha-\beta)\delta\right) \end{cases}\tag{*2}$$ Furthermore, if one substitute any integers $\lambda, \alpha,\beta,\gamma,\delta$ into $(*2)$ and if the resulting $a, b, c, d$ are integers, then it will be a solution of $(*1)$. It is not hard to check this happens when and only when

$$\lambda(\alpha+\beta+\gamma+\delta)\quad\text{ is an even number }\tag{*3}$$

Conclusion - all solutions of $(*1)$ can be parametrized by $(*2)$ subject to the constraint $(*3)$.

The formula you have is a special case of above parametrization. It can be reproduced by following substitutions:

$$\begin{cases} \lambda &= 1\\ \alpha &= s - p,\\ \beta &= 2(s+p+k),\\ \gamma &= k+p,\\ \delta &= k+s. \end{cases}$$

• It is not clear from (1) we obtain the following equation? The existence of Pythagorean quadruples has no relation to the solution of the form: $R(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$ I double-checked, but your formula can not find the solution of equation (1). – individ Mar 28 '14 at 12:12
• @individ If you plug the formula of $\alpha,\ldots,\delta$ in terms of $k,s,p$ I wrote down above into $(*2)$ and ask a CAS to simplify it, you will get the 4 polynomials you write down in the beginning of your post. – achille hui Mar 28 '14 at 12:45
• I did so, but it is impossible. These solutions are not solutions of (1) – individ Mar 28 '14 at 12:57
• I don't understand what you mean it is impossible. I randomly pick $(\alpha,\beta,\gamma,\delta)$ as $(1,3,5,7)$ and plug it into $(*2)$. It give me $(a,b,c,d) = (18,8,56,2)$ and $2(18^2+8^2+56^2+2^2)$ does equal to $(18+8+56+2)^2$. – achille hui Mar 28 '14 at 13:06
• you $(r,q,e,t)=(1,3,5,7)$ According to your formula should be $t-e=q=s-p$ , $t-e=7-5=2$ , $q=1$. Better rewrite in another form, these letters are not comfortable to type. – individ Mar 28 '14 at 13:35

EDITTTT: two research level articles on this, one by Kontorovich and one by Fuchs, can be downloaded for free at AMS BULLETIN APRIL 2013. There is also a short survey article by Peter Sarnak, in the April 2011 MAA Monthly. I have a pdf of that, if anyone is interested.

Alright, I see part of what is going on. The standard recipe, stereographic projection around a given integral solution, is guaranteed to give all rational solutions; this method parametrizes solutions with four components by three parameters, or generally $k$ components by $k-1$ parameters. And, some choices of central point for projection give more efficient recipes than others; the one individ chose was very good.

However, it appears that such a recipe is not going to give all integral solutions, nor will a finite number of such recipes. So, here is my version, with rational solutions. That is, I took coprime $(x,y,z)$ and calculated

$$a = x^2 + y^2 + 2 z^2 - y z - z x - 2 x y,$$

$$b = x^2 - y z + z x,$$

$$c = y^2 + y z - z x,$$

$$d = 2 z^2 + y z + z x.$$

Then I took $$g = \gcd(a,b,c,d),$$ and divided all four of $(a,b,c,d)$ by that. As a result, I found all the "root" solutions given at TABLE. I am encouraged that $g$ was always the sum of two squares. Hmmm; actually, it is easy to show that $\gcd(x,y,z) = 1$ implies that $g = \gcd(a,b,c,d)$ is not divisible by $4$ or by any prime $q \equiv 3 \pmod 4,$ so we can always write $g = u^2 + v^2$ with integers and $\gcd(u,v)=1.$

Oh, root solutions have $$a \leq 0 \leq b \leq c \leq d,$$ $$a+b+c+d > 0,$$ $$a + b + c \geq d.$$ These were defined in Graham, Lagarias, Mallows, Wilks, Yan, Apollonian circle packings: Number Theory, Journal of Number Theory, volume 100 (2003) pages 1-45. It was shown that every integral solution can be connected to such a root solution by Vieta jumps, thus dividing the solutions into a forest of countably many rooted trees. An unusual feature is that these Vieta jumps are (they must be) elements in a group of $4$ by $4$ invertible integral matrices, called the Apollonian Group, which is just the (orthogonal or rotation or automorphism) group for the quadratic form/indefinite lattice given by $$a^2 + b^2 + c^2 + d^2 -2ab-2ac-2ad-2bc-2bd-2cd,$$ with Gram matrix $$H \; = \; \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ -1 & 1 & -1 & -1 \\ -1 & -1 & 1 & -1 \\ -1 & -1 & -1 & 1 \end{array} \right).$$ As a matrix, eigenvalues are $-2,2,2,2$ with orthogonal (but not orthonormal) eigenvectors as columns of $$W \; = \; \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 2 & -1 \\ 1 & 0 & 0 & 3 \end{array} \right).$$

   a       b       c       d             g       x       y       z
-1       2       2       3   gcd was   1       1       0       1
-2       3       6       7   gcd was   5      -3      -6       2
-3       4      12      13   gcd was  10       4      12      -3
-3       5       8       8   gcd was   1       2       3      -1
-4       5      20      21   gcd was   5      -2       1      -7
-4       8       9       9   gcd was  29      -1      12       9
-5       6      30      31   gcd was  26      -6     -30       5
-5       7      18      18   gcd was  10      -4       2      -9
-6       7      42      43   gcd was  37      -7     -42       6
-6      10      15      19   gcd was  13     -10     -15       6
-6      11      14      15   gcd was   2       0      -4      -3
-7       8      56      57   gcd was  13      -3       2     -19
-7       9      32      32   gcd was  17       5      -3      16
-7      12      17      20   gcd was   5      -8      -9      -4
-8       9      72      73   gcd was  65       9      72      -8
-8      12      25      25   gcd was  25     -14     -27       8
-8      13      21      24   gcd was   5       7      11      -4
-9      10      90      91   gcd was  82     -10     -90       9
-9      11      50      50   gcd was   5      -4     -17       3
-9      14      26      27   gcd was  26     -16     -28       9
-9      18      19      22   gcd was  17      -1      13      11
-10      11     110     111   gcd was  25      -4       3     -37
-10      14      35      39   gcd was  10      -6       2     -13
-10      18      23      27   gcd was   1      -4      -5       2
-11      12     132     133   gcd was 122     -12    -132      11
-11      13      72      72   gcd was  37      -7       5     -36
-11      16      36      37   gcd was  34     -18     -38      11
-11      21      24      28   gcd was   5      -7       1      -7
-12      13     156     157   gcd was 145     -13    -156      12
-12      16      49      49   gcd was   5      -6     -17       4
-12      17      41      44   gcd was   1      -3      -7       2
-12      21      28      37   gcd was  25      21      28     -12
-12      21      29      32   gcd was   2      -6      -8       3
-12      25      25      28   gcd was   1      -5      -5      -2
-13      14     182     183   gcd was  41      -5       4     -61
-13      15      98      98   gcd was  50      -8       6     -49
-13      18      47      50   gcd was   2      -2      -8      -5
-13      23      30      38   gcd was  26     -16       2     -19
-14      15     210     211   gcd was 197     -15    -210      14
-14      18      63      67   gcd was  53     -18     -63      14
-14      19      54      55   gcd was   1      -2       1      -5
-14      22      39      43   gcd was  37     -24     -41      14
-14      27      31      34   gcd was  10     -16     -18       7
-15      16     240     241   gcd was 226     -16    -240      15
-15      17     128     128   gcd was  13      -6     -43       5
-15      24      40      49   gcd was  34     -24     -40      15
-15      24      41      44   gcd was  10       2      16      11
-15      28      33      40   gcd was   2       0      -6      -5
-15      32      32      33   gcd was   2       8       8      -3
a       b       c       d             g       x       y       z


• +1 even though I don't fully understand this. No idea what Vieta jumps are but I know Apollonian group is a finitely presented group generated by 4 involutions. Do the Vieta jumps correspond to those 4 generators? (or actions of them???) Just try to have a rough feeling of what happens. – achille hui Mar 30 '14 at 1:43
• @achillehui, yes, the four involutions are simply jumps in each of the four variables. The original example is the tree of solutions to the Markov equation, although that is not homogeneous en.wikipedia.org/wiki/Markov_number Anyway, given an Apollonian solution $(a,b,c,d),$ you can get another solution by changing just one out of four, keeping the other three fixed. Same on the Markov tree, you move around by changing one number at a time. – Will Jagy Mar 30 '14 at 1:55
• @achillehui, I just use the term Vieta jumping because it is a popular technique for contest problems, so known to some on this site. Markov: if I have a particular solution $(a,b,c)$ to $$x^2 + y^2 + z^2 = 3 x y z,$$ I get three neighboring solutions as $(3bc-a,b,c)$ and $(a,3ac-b ,c)$ and $(a,b,3ab-c).$ An Apollonian quadruple $(a,b,c,d)$ has four neighbor solutions. – Will Jagy Mar 30 '14 at 2:19
• I see, learns something new everyday. Thanks. – achille hui Mar 30 '14 at 3:17
• @achillehui, given Apollonian solution $$(a,b,c,d),$$ the four neighboring solutions are $$(2b+2c+2d-a,b,c,d),$$ $$(a,2a+2c+2d-b,c,d),$$ $$(a,b,2a+2b+2d-c,d),$$ $$(a,b,c,2a+2b+2c-d).$$ – Will Jagy Mar 30 '14 at 3:27

The above equation which is mentioned below has another parametrisation,

$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$

$a=(5k+2)^2$

$b=(3k-4)^2$

$c=(2k+6)^2$

$d=4(19k^2+10k+28)$

For $k=0$ we get $(a,b,c,d)=(1,4,9,28)$

• A bad parameterization. Because that describes a very small number of solutions. – individ Jul 7 '17 at 7:57

You can consider another equation:

$2(a^2+y^2+c^2+d^2+u^2)=(a+y+c+d+u)^2$

And write the formula to solve this equation.

$a=-(k^2+2(q+t+b)k+b^2+q^2+t^2+bq+bt+qt)$

$y=k^2+2(q+t+b)k+2b^2+q^2+t^2+2bq+2bt+qt$

$c=k^2+2(q+t+b)k+b^2+2q^2+t^2+2bq+bt+2qt$

$d=k^2+2(q+t+b)k+b^2+q^2+2t^2+bq+2bt+2qt$

$u=2k^2+2(q+t+b)k+b^2+q^2+t^2$

$k,q,t,b$ - what some integers.

My more general method of calculation. Enables us to solve and other factors. Find out whether or when given coefficients solutions and immediately write the formula.

For example, consider the equation:

$4(a^2+b^2+c^2+d^2)=3(a+b+c+d)^2$

Then the solutions are of the form:

$a=-(p^2+4(k+s)p+2k^2+2ks+2s^2)$

$b=p^2+4(k+s)p+6k^2+6ks+2s^2$

$c=p^2+4(k+s)p+2k^2+6ks+6s^2$

$d=3p^2+4(k+s)p+2k^2-2ks+2s^2$

I think the best result by a direct solution of these equations. Brute force search of the law or does not make sense. Spend a lot of effort, but the result does not always work.