The conjecture is *false*, and in fact there exists a positive integer solution for
$$\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}=5,$$
though I was unable to find it explicitly. I will explain how to prove it, though I remark that my method is no different that the one of Bremner and MacLeod.

Take the following SageMath code:

```
n = 5
F = a*(2*a+3*b)*(2*c+3*a)+b*(2*a+3*b)*(2*b+3*c)+c*(2*b+3*c)*(2*c+3*a) - n*(2*a+3*b)*(2*b+3*c)*(2*c+3*a)
O = [3,-2,0]
E = EllipticCurve_from_cubic(F,O,morphism = False)
f = EllipticCurve_from_cubic(F,O,morphism = True)
finv = f.inverse()
```

$F$ is the (homogeneous) cubic polynomial whose roots are the solutions to the equation above, which we treat as a projective curve, and $O$ is an obvious point on this cubic (note that it corresponds to one of the denominators vanishing, so doesn't give a solution to the original equation). $E$ is an elliptic curve in the Weierstrass form resulting from transforming $F$ and $O$, and $f$ is an algebraic isomorphism from $F$ to $E$. $finv$ is the inverse of $f$. Whatever $f$ and $f^{-1}$ are, they give a bijection between rational points on the two curves.

$\newcommand{\Q}{\mathbb Q}$Our goal is to prove that there is a point $P_0\in E(\Q)$ such that the coefficients of $f^{-1}(E)$ are positive. To generate many points in $E(\Q)$, we first have to find some, which is done in SageMath using the command

```
E.gens()
```

which returns some points of infinite order in $E(\Q)$ (contrary to the name of the command, the points are not necessarily guaranteed to generate the entire group). In this case, it returns the following pair of points:

```
(-9688590530211424458781697369983/59430279362573596065056333378 : -109157965923796779917012462777322533033259926157/128057987622979511510369081256538361341797175 : 1)
(-3102338504770020903627975887607/29715139681286798032528166689 : -8926749292636238116520391951005662916568902402171/27660525326563574486239721551412286049828189800 : 1)
```

(if you wish to repeat this computation, be warned it took over 10 minutes.) Let $P$ be the first of those points. The crucial thing about this point can be seen when we make a plot:

```
E.plot()+plot(P)
```

We see that $P$ lies on the *egg* of $E$, the bounded connected component of $E(\mathbb R)$ (it is impossible to see in the picture, but the "self-crossing" of $E$ is actually a place where $E$ splits into two components). It is classical that for $E$ like this, with disconnected set of real points, we have
$$E(\mathbb R)\cong(\mathbb R/\mathbb Z)\times\mathbb Z/2\mathbb Z$$
as a group, and a point $P$ lying on the egg corresponds to a point $(\alpha,1)$ for some irrational $\alpha$. For this reason we see that the subgroup of $E(\Q)$ generated by $P$ is dense in $E(\mathbb R)$.

Now we make our final observation: our original equation has a real positive solution, so points of $E(\mathbb R)$ corresponding to solutions with coefficients of the same sign form a nonempty open subset. Therefore some multiple $kP$ of $P$ belongs to this open subset, and the point $f^{-1}(kP)\in F(\Q)$ will have positive coordinates (or, all negative, but then we can just flip the signs). Clearing the denominators we find an integral solution to $F=0$.

To find an actual solution we can try computing $kP$ until $f^{-1}(kP)$ has positive coordinates, which the following code does:

```
for k in range(3,1000):
Q = finv(k*P)
if R[0]>0 and R[1]>0:
k,R
```

Unfortunately, running this for a while I found no hits. The region of positive solutions is rather small (compare with the figures in Bremner-MacLeod; I didn't compute how small they are here) so the least such $k$ could be quite large. I think $k$ got to around $200$ before I killed the program, which heuristically means that the least solution we get this way will have on the order of $10000$ digits, though I make no claims that this is the smallest possible solution (we can probably do much better by taking linear combinations of $P$ and the second point given by `E.gens()`

, but I didn't bother to check those).

This resolves the case $n=5$. For $n=1,3$ none of the generated points (one for $n=1$, two for $n=3$) lie on the egg, so the method does not guarantee existence of solutions. However, it doesn't exclude them either: for one, `E.gens()`

does not guarantee we have generators of $E(\Q)$, and further I do not know if the positivity region lies entirely within the egg. Both problems are surmountable though: I believe Sage has a method to compute generators with certainty, and we can check whether the positivity region lies on the egg by plotting points of (the affine part of) $F$. I have made no attempts to try larger values of $n$, but feel free to play around with the code yourself.