Odd abundant numbers and the condition $\sqrt{n^2+12n\sigma(n)}\in\mathbb{Z}_{\geq 1}$

We denote for integers $n\geq 1$ the sum of divisors function as $\sigma(n)=\sum_{d\mid n}d$.

If we assume that $m$ is an odd perfect number then it satisfies that $m^2+12m\sigma(m)$ is a perfect square, that is $$\sqrt{m^2+12m\sigma(m)}=\sqrt{m^2+24m^2}$$ is an integer.

The existence of odd perfect numbers is an open problem.

On the other hand an integer $n\geq 1$ is deficient if $\sigma(n)-2n<0$, as you can see from this Wikpedia dedicated to Deficient number.

Question. Prove or refute:

If an odd integer $n\geq 1$ satisfies that $$\sqrt{n^2+12n\sigma(n)}$$ is integer, then our odd integer $n$ is deficient or perfect.

That I mean (since the existence of odd perfect numbers is an open problem) and the issue in which I'm am interested is in the veracity of next/previous conjecture:

There are no odd abundant numbers $n$ for which $\sqrt{n^2+12n\sigma(n)}$ is integer.

Many thanks.

I've tested it upto $10^7$. Thus I presume that our work is to study the equation $$12n\sigma(n)=k^2-n^2,$$ for (those integers being solutions of our previous condition) positive integers $k$, and elucidate if we can rule out our odd integer $n$ as an abundant number (this is the related Wikipedia Abundant number). The RHS of previous equation has factorization, thus we write $$12n\sigma(n)=(k-n)(k+n).$$

Because maybe there exist a counterexample, then please provide the first odd abundant numbers satisfying the condition in the conjecture (if you are able to find more examples provide the first few in a table) and provide the explicit calculations showing that yours is a legitimate counterexample.

The evidence that you have provided for the truth of this conjecture is actually quite weak. There are only about $20000$ odd abundant integers up to $10^7$ (the smallest of course being $945$). For each such value, if we were to just choose an random integer about the same size as $n^2+12n\sigma(n)$, then it would have about a $1/(2\sqrt{n^2+12n\sigma(n)})$ chance of being a square. When $n$ is abundant this quantity is at most $\frac{1}{10n}$.

When you sum up these probabilities across the $20000$ examples you tested, the expected number of squares found is less than $0.002$. So even if the conjecture was false and there is no theorem forcing these numbers to be non-square, I would consider it to be rather lucky if you had found an example below $10^7$. It turns out $10^7$ was actually not a very wide net to catch if you hoped to find an example.

Heuristically, one expects there to be infinitely many examples, as the odd abundant integers have positive density (though it is not known to very high precision, I believe it is known to be at most $0.08$) and the sum $\sum_{n=1}^\infty 1/(2\sqrt{n^2+12n\sigma(n)})$ is divergent since $\sigma(n) = O(n \log \log n)$.

However, the rate of divergence is so slow that unless we're quite lucky it is probably unreasonable to demand the smallest example: as a very crude guess, it might be as large as $10^{32}$. (I arrived at this guess by assuming the odd abundant numbers starting from $10^7$ are evenly distributed with density $1/12$ and used the more generous upper bound of $\frac{1}{10n}$ for the expected value).

• Your reasoning has a great mathematical content. Of couse I accept your words about the computational evidence but I did not know the theorems and heuristics that you told us. In summary, your answer is perfect.
– user243301
Jun 3, 2018 at 9:06
• As it turns out, the smallest counterexample actually is barely above the limit tested by @user243301: $n=14259375=3^{3}\times5^{5}\times13^{2}$ is abundant, since $\sigma(n)=28591920=2n+73170$ and $\sqrt{n^2+12n\sigma(n)}=71384625$. Jun 3, 2018 at 15:54
• @PeterKošinár Your comment sounds like an excellent answer to me :). Jun 3, 2018 at 17:04
• Many thanks @PeterKošinár for your counterexample.
– user243301
Jun 3, 2018 at 17:08
• 71384625 = 3^3 x 5^3 x 13 x (5^3 x 13 + 2) Jun 6, 2018 at 17:33