# Find all the natural solutions of (a+b+c)a-3bc=0

I was trying to solve a geometry puzzle when I came across a simple algebraic problem that I couldn't solve.

Given the expression $(a+b+c)a - 3bc = 0$, find all natural solutions for $a$, $b$ and $c$.

I've tried to isolate one variable, like $$a=\frac{-b-c\pm\sqrt{b^2+14bc +c^2}}{2}$$ but I didn't get anywhere. Despite this, I've notice that the numbers $(2, 6, 1)$ satisfy the condition. Does anyone can help with this problem model?

• As an aside, you could continue from there by noting that, for $a$ to be natural, at the very least you must have that $b^2 + 14bc + c^2$ is a perfect square, so one could solve $b^2 + 14bc + c^2 = d^2$ similarly to how the Pythagorean equation is solved. – user14972 Sep 29 '17 at 2:02
• @AmateurMathGuy notice that here all edits are welcome, with the purpose to improve the formatting (for example yours wiht the second formula). In my edit of this post (I've edited few questions, thus my experience isn't the best) I respect the choice of tags but I think that it was required add the tag (diophantine-equations), also notice that in the title (a+b+c)a-3bc=0 doesn't refer a string of letters, because it is an equation. Many thanks, and good day. – user243301 Sep 29 '17 at 10:12
• @user if you're referring to the chat, I was talking about other, more subjective edits. Thank you – AmateurMathPirate Sep 29 '17 at 10:43

You might like this better, as it is just solving your $$w = \sqrt {b^2 + 14 bc + c^2}$$ an integer.

We take $x \geq 0,$ $\gcd(x,y) = 1,$ and these two recipes only: $$\mbox{I:} \; \; w = |11 x^2 + 2 xy - y^2|, \; \; \; b = 6 x^2 + 5xy + y^2, \; \; \; c = x^2 - xy,$$

$$\mbox{II:} \; \; w = |44 x^2 + 20 xy + 2 y^2|, \; \; \; b = 21 x^2 + 13xy + 2y^2, \; \; \; c = 5x^2 + xy,$$ and discard triples when either $b < 0$ or $c < 0$ or $\gcd(w,b,c) \neq 1.$

=====================================

  w      b      c               x      y
1      1      0     I         0      1
4      1      1    I I        1     -4
11      6      1     I         1      0
13      2      5     I         1     -4
23      3     10     I         2     -3
37     12      7     I         1     -6
44     21      5    I I        1      0
44      5     21    I I        3     -8
47     35      2     I         2      1
52      3     35    I I        5    -18
52     35      3    I I        3    -14
59     10     21     I         3     -4
61      4     39     I         3    -10
71      5     44     I         4     -7
73     15     22     I         2     -9
83     28     15     I         3     -2
92     55      7    I I        1      2
92      7     55    I I        5    -14
97      5     68     I         4    -13
107     88      3     I         3      2
109     56     11     I         1    -10
121     35     26     I         2    -11
131     14     65     I         5     -8
143     45     28     I         4     -3
143      7    102     I         6    -11
148    117      5    I I        5    -24
148      5    117    I I        9    -32
157     90     13     I         1    -12
167     77     20     I         4     -1
169     21     76     I         4    -15
179     36     55     I         5     -6
181     40     51     I         3    -14
188     33     65    I I        5    -12
188     65     33    I I        3     -4
191    165      4     I         4      3
193      7    150     I         6    -19
227     18    133     I         7    -12
229     24    115     I         5    -18
236     11    171    I I        9    -26
236    171     11    I I        1      6
239      9    184     I         8    -15
241     99     34     I         2    -15
244     35     99    I I        9    -34
244     99     35    I I        7    -30
251    104     35     I         5     -2
253     70     57     I         3    -16
253      8    203     I         7    -22
263     55     78     I         6     -7
277    182     17     I         1    -16
284    119     39    I I        3     -2
284     39    119    I I        7    -18
292    247      7    I I        7    -34
292      7    247    I I       13    -46
299    266      5     I         5      4
299     44    119     I         7    -10
311     91     66     I         6     -5
313    143     38     I         2    -17
332     13    253    I I       11    -32
332    253     13    I I        1      8
337     77     92     I         4    -19
347     22    225     I         9    -16
349    240     19     I         1    -18


=====================================

MOST solutions have $a=b=c.$ However, many others. Also, given solution $$(a,b,c),$$ $$(a,c,b)$$ is another solution, which seems distinct as far as my four recipes. The two orders will both appear, usually one order with $|x| + |y|$ pretty small, but the other order with $|x| + |y|$ larger.

In these FOUR recipes, take $\gcd(x,y) = 1,$ and if $a < 0$ or if $\gcd(a,b,c) \neq 1,$ discard that triple. $$\mbox{I:} \; \; \; a = 3 xy, \; \; b = xy + 2 y^2, \; \; c = 2 x^2 + xy$$ $$\mbox{II:} \; \; \; a = 2 x^2 + xy - y^2, \; \; b = 2 x^2 -xy , \; \; c = 2 x^2 + 3xy + y^2$$ $$\mbox{III:} \; \; \; a = 3 x^2 -3 xy, \; \; b = 5 x^2 -xy , \; \; c = 2 x^2 - 3xy + y^2$$ $$\mbox{IV:} \; \; \; a = 7 x^2 -3 xy , \; \; b = 35 x^2 -29xy + 6 y^2 , \; \; c = 3 x^2 - xy$$

The outcome of the Fricke-Klein method is that there need be only a finite set of such recipes. You may take just one recipe if you are willing to then divide out by $\gcd(a,b,c),$ but then you do not now how big to take the variables.

All coprime positive solutions other than $1,1,1,$ with $a \leq 100$

  2      6      1
3      5      2
5     10      3
5     35      2
7     35      3
8     88      3
9     12      7
9     21      5
9     39      4
11     44      5
12     68      5
14     21     10
15     55      7
18     22     15
20     28     15
21     56     11
26     65     14
27     90     13
30     35     26
35     45     28
35     77     20
36     76     21
44     55     36
45     51     40
45     65     33
54     99     34
55     99     35
63     70     57
65     78     55
77     91     66
84     92     77


=========================================

using the four recipes

  2      1      6     I I        1      1
3      5      2  I I  I        1      0
5     10      3     I I        2     -1
5      2     35     I I        2      3
7      3     35  I  V        7     16
7     35      3  I  V        1      0
8      3     88     I I        3      5
9     21      5     I        1      3
9     39      4  I I  I        3      2
9      5     21     I        3      1
9      7     12  I I  I        1     -2
11     44      5     I I        4     -3
12     68      5  I I  I        4      3
14     21     10     I I        3     -1
15     55      7     I        1      5
15      7     55     I        5      1
18     22     15  I I  I        2     -1
20     15     28     I I        3      1
21     11     56  I I  I        1     -6
26     65     14     I I        5     -3
27     13     90  I I  I        1     -8
30     26     35  I I  I        2     -3
35     20     77     I I        4      3
35     28     45     I I        4      1
36     76     21  I I  I        4      1
44     55     36     I I        5     -1
45     33     65     I        5      3
45     51     40  I I  I        3     -2
45     65     33     I        3      5
54     34     99  I I  I        2     -7
55     35     99  I  V       11     24
55     99     35  I  V        5      8
63     57     70  I I  I        3     -4
65     78     55     I I        6     -1
77     66     91     I I        6      1
84     92     77  I I  I        4     -3


============================

Rewrite the equation as $(a+b)(a+c)=4bc$. Now \begin{eqnarray*} a+b= 2 \alpha \beta \\ a+c = 2 \gamma \delta \\ b= \alpha \gamma \\ c= \beta \delta \end{eqnarray*} will satisfy this provided $\alpha ( 2 \beta - \gamma)=a= \delta(2 \gamma - \beta)$. So choose \begin{eqnarray*} \alpha=2 \gamma - \beta \\ \delta= 2 \beta - \gamma \end{eqnarray*} and so we have faimily of solutions generated by \begin{eqnarray*} a= (2 \gamma - \beta)( 2 \beta - \gamma) \\ b= \gamma (2 \gamma - \beta)\\ c= \beta ( 2 \beta - \gamma).\\ \end{eqnarray*}

$$c=\frac{a^2+ab}{3b-a}$$

The problem then becomes: for which $a,b$ do we have $3b-a|a(a+b)$.

Let $d= gcd(a,b)$ then $a=da', b=db'$ with $gcd(a',b')=1$. Then

$$3b'-a'|da'(a'+b')$$

Now, let us observe that $$gcd(3b'-a', a')|3gcd(a',b')=3$$ and $$gcd(3b'-a', a'+b')|4gcd(a',b')=4 \,.$$

At this point the problem reduces to 6 cases, and a case by case analysis will complete the problem: $$gcd(3b'-a', a') \in \{1,3\} \, \mbox{ and } \, gcd(3b'-a', a'+b') \in \{ 1,2,4\}$$

But one can also write the general solution this way: Pick $a',b'$ arbitrary. Let $$d=\frac{3b'-a'}{gcd(3b'-a', a')gcd(3b'-a', a'+b')} \cdot l$$ for some integer $l$.

Then $$a=da'=\frac{3b'-a'}{gcd(3b'-a', a')gcd(3b'-a', a'+b')} \cdot la' , \\ b=\frac{3b'-a'}{gcd(3b'-a', a')gcd(3b'-a', a'+b')} \cdot l b', \\ c=\frac{a^2+ab}{3b-a}= \frac{a'(a'+b')}{gcd(3b'-a', a')gcd(3b'-a', a'+b')} \cdot l$$ is a solution, and this describes the general solution.