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I would like to find the solutions to $$i)\qquad a(a+b+c)=bc \\ii)\qquad a(a+b+c)=2bc \\iii)\qquad a(a+b+c)=3bc$$ for $0< a \le b \le c$ and of course: $\textrm{gcd}(a,b,c) = 1$ (since those are the interesting cases).

To be honest, Diophantine equations seem like a black art. Guess and check has always been the only sure method that I can understand.

At least the last case ($iii$) seems to be "easy" as the only solution is apparently $ a = b= c=1$. I don't see how to prove this though.

For $i$, I've found solutions such as: $\{\{1,2,3\}, \{2,3,10\}, \{3,4,21\}, \{3,5,12\}, \ldots\}$

And For $ii$, I've found solutions such as: $\{\{1,1,2\}, \{4,5,6\}, \{15,20,21\}, \{20,22,35\}, \ldots\}$

Notwithstanding my ability to exhaust over integers, I don't have any deeper understanding of what is going on here mathematically. I would really like to understand these better and would appreciate some help! Thanks in advance!

amcalde
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  • these are indefinite ternary quadratic forms (put everything one one side, set equal to zero); if there are any solutions in integers, there are infinitely many. The inequalities are not a natural part of the setup, and may reduce to finitely many solutions. – Will Jagy Dec 10 '14 at 23:34
  • I guess the inequalities are not really needed. I'll search on "indefinite ternary quadratic forms." That is a mouthful. – amcalde Dec 10 '14 at 23:35
  • The diophantine equation $a(a+b+c)=nbc$ is equivalent to $$a^2(1/n+1/n^2)=(b-a/n)(c-a/n).$$ – Ian Mateus Dec 11 '14 at 00:38
  • It's quite easy to give a parametric solution to $a(a+b+c)=nbc$, but the tricky part is the OP wants $0<a<b<c$ which restricts the solution space. I'll give a partial answer below. – Tito Piezas III Dec 11 '14 at 00:54

1 Answers1

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One can give a parameterization to,

$$a(a+b+c) = nbc\tag{1}$$

However, since you want $0<a<b<c$, this makes it trickier.

For $n=1$: A look at your solutions and one can see that $a,b$ are in arithmetic progression. Hence,

$$a,\;b,\;c = n,\;n+1,\;2n^2+n$$

For $n=2$: Most of your $b,c$ differ by one. Hence,

$$a,\;b,\;c = 2pq,\;p^2+pq-1,\;p^2+pq$$

where $p,q$ solve the Pell equation $p^2-3q^2=1$. Since $pq<p^2$, this guarantees that $a<b$.

For $n\ge3$: I checked this with Mathematica and, curiously, there doesn't seem to be any.

P.S. Without the inequalities, then $(1)$ has a two-parameter solution,

$$a,\;b,\;c = (-u + n v)u, \; (u + v)u, \; (-u + n v)v$$

Edit: This answers the comment below. First, we do the substitution $b,\;c = x+y,\;x-y\,$, on $(1)$ then solve for $a$,

$$a =-x\pm\sqrt{(n+1)x^2-ny^2}$$

Thus we have to solve, as W. Jagy pointed out, a ternary quadratic form,

$$(n+1)x^2-ny^2=z^2$$

and initial solution of which is $x,y,z =1,1,1$. Given an initial solution, we generally can find an infinite more. Two methods are described here in Form 2b.

Tito Piezas III
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