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?
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
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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.
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 $\gcd(w,b,c) \neq 1.$
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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
=====================================