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Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false. Note that $a+b$ must be odd. Put $\ell = a+b$ (and $\ell$ is odd) and $m=a-b$ which is also odd. The condition that $b<a<2b$ now translates to $0<m < \ell/3$. The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.

So the question really is can $m^3-2$ have a square factor of size exceeding $(3m)^2$? Here's a heuristic that indicates why this should be possible. Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$. Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev). For $p=3$ one should work a tiny bit harder. Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$. We may expect that probability to be about $f(\ell)/(3\ell)$. Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m< \ell/3$) should be about $$ \sum_{\ell < x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x, $$ for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$. This suggests that there are infinitely many counterexamples.

I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.

Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false. Note that $a+b$ must be odd. Put $\ell = a+b$ (and $\ell$ is odd) and $m=a-b$ which is also odd. The condition that $b<a<2b$ now translates to $0<m < \ell/3$. The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.

So the question really is can $m^3-2$ have a square factor of size exceeding $(3m)^2$? Here's a heuristic that indicates why this should be possible. Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$. Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev). For $p=3$ one should work a tiny bit harder. Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$. We may expect that probability to be about $f(\ell)/(3\ell)$. Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m< \ell/3$) should be about $$ \sum_{\ell < x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x, $$ for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$. This suggests that there are infinitely many counterexamples.

I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.

Added: A relevant paper for such problems is Granville's paper in IMRN.
Granville shows that on the abc conjecture the largest square factor of any cubic polynomial (without repeated roots) $f(m)$ is at most $m^{2+o(1)}$. This generalizes the observation in GH from MO's answer (the general case requires more work than the easy $x^3-2$ example). Granville also gives reasons to believe that this exponent is best possible, and in particular shows that such large square factors are attained for cubic polynomials having a rational root -- this uses the solutions to Pell's equation. Of course, Granville's construction doesn't apply for $x^3-2$ which is irreducible, but the results may be suggestive. Also note that the Pell type constructions produce a logarithmic number of examples with large square factors, which is again suggestive.

Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false. Note that $a+b$ must be odd. Put $\ell = a+b$ (and $\ell$ is odd) and $m=a-b$ which is also odd. The condition that $b<a<2b$ now translates to $0<m < \ell/3$. The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.

So the question really is can $m^3-2$ have a square factor of size at least $(3m)^2$? Here's a heuristic that indicates why this should be possible. Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$. Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev). For $p=3$ one should work a tiny bit harder. Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$. We may expect that probability to be about $f(\ell)/(3\ell)$. Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m< \ell/3$) should be about $$ \sum_{\ell \le x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x, $$ for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$. This suggests that there are infinitely many counterexamples.

I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.

Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false. Note that $a+b$ must be odd. Put $\ell = a+b$ (and $\ell$ is odd) and $m=a-b$ which is also odd. The condition that $b<a<2b$ now translates to $0<m < \ell/3$. The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.

So the question really is can $m^3-2$ have a square factor of size exceeding $(3m)^2$? Here's a heuristic that indicates why this should be possible. Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$. Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev). For $p=3$ one should work a tiny bit harder. Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$. We may expect that probability to be about $f(\ell)/(3\ell)$. Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m< \ell/3$) should be about $$ \sum_{\ell < x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x, $$ for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$. This suggests that there are infinitely many counterexamples.

I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.

Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false. Note that $a+b$ must be odd. Put $\ell = a+b$ (and $\ell$ is odd) and $m=a-b$ which is also odd. The condition that $b<a<2b$ now translates to $0<m \le \ell/3$. The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.

So the question really is can $m^3-2$ have a square factor of size at least $(3m)^2$? Here's a heuristic that indicates why this should be possible. Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$. Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev). For $p=3$ one should work a tiny bit harder. Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$. We may expect that probability to be about $f(\ell)/(3\ell)$. Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m\le \ell/3$) should be about $$ \sum_{\ell \le x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x, $$ for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$. This suggests that there are infinitely many counterexamples.

I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.

Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false. Note that $a+b$ must be odd. Put $\ell = a+b$ (and $\ell$ is odd) and $m=a-b$ which is also odd. The condition that $b<a<2b$ now translates to $0<m < \ell/3$. The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.

So the question really is can $m^3-2$ have a square factor of size at least $(3m)^2$? Here's a heuristic that indicates why this should be possible. Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$. Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev). For $p=3$ one should work a tiny bit harder. Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$. We may expect that probability to be about $f(\ell)/(3\ell)$. Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m< \ell/3$) should be about $$ \sum_{\ell \le x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x, $$ for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$. This suggests that there are infinitely many counterexamples.

I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.