$24ml+1=k^2$ has no solution for all $l=1 \dots m$

Question

Investigating solutions of $$24ml+1=k^2$$ for $l=1\dots m$

The question is to find the $m$-s for which the above equation has no solution for all $l=1..m$-s. The first few $m$-s are: $$3, 9, 24, 27, 81, 192, 243, 432, 729$$ Actually have found that the $m$ should be of form $2^a3^b$. Seems hang on the simple task, of finding general formula for this.

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One interesting addition. Just tested the cases when the $$12ml+1=k^2$$ has no solution for all $$l=1..m$$ Seems it has no solution iff $m=3^a$ But has no proof yet.

Answer 1

Under $100$ there are $84$ $m$-values yield no integer integers $(k)$ for $l=1$.

To find them is simple. $\quad k\not\in\mathbb{N} \text { for }k=\sqrt{24 m + 1)}$.

For example, for any non-zero integer $(m)$, $$\text{If }\space \sqrt{24 m + 1)}\ne \big\lfloor \sqrt{24 m + 1)}\space\big\rfloor \text{, then print } m \text{, endif }\tag{1}$$

This is the same as testing $\quad\sqrt{24 m \times(1) + 1)},\quad $ i.e. $l=1.\quad$ To find non-solution m-values for any other l-values, you must test the full $\quad 24m\times(l)+1, l\le m.\quad$ For examples,

\begin{align*} \text{If }l=2,\quad& m\in\{ \emptyset \}\\ \text{If }l=3,\quad& m\in\{ 1,2,3\}\\ \text{If }l\in\{4,5,6,7,8\}\quad& m\in\{ \emptyset \}\\ \text{If }l=9,\quad& m\in\{ 1,2,3,4,5,6,7,8,9\}\\ \text{If }l\in\{10,11,12,\cdots, 23\}\quad& m\in\{ \emptyset \}\\ \mathbf{\vdots} \end{align*} Your derived list is correct but I cannot offer a better formula for finding these m-values.

$\textbf{Update}\\$ We can solve for $l$ and see which k-values yield an integer for a given m-value.

$$l = \frac{k^2 - 1}{24 m} \quad\text{and we know }m\ne0\tag{2}$$

We can find $k^2$ by taking multiples of $24m$ and adding 1 to see if any $\sqrt{k^2}$ yields an integer. Having found $k$ we plug it back into equation $(2)$ and find $l$. We have a valid $m$ if and only if the smallest $l$ generated is greater than $m$. The test values for $m$ can quickly be obtained from equation $(1)$ above.

Answer 2

Here are a couple of results that can be proved:

There is no solution if $m=3^a, a>0.$

For $24ml=(k-1)(k+1)$ we must have $k=6x\pm 1$. Then$$ml=\frac {x(3x\pm 1)}{2}.$$

Now if $m=3^a, a>0$, then $m$ has to be a factor of $x$. But then $l\ge \frac {3x- 1}{2}>m,$ a contradiction.

There is a solution if $m$ is not divisible by $3$.

Let $2m=3l\pm 1$. Then $l\le m$ and $$24lm+1=12l(3l\pm 1)+1=(6l\pm 1)^2.$$

Answer 3

The OP also conjectures that $12ml+1=k^2$ has no solutions for $1\le l\le m$ if and only if $ m=3^a$. This is true and can be proved as follows.

Only if

For $12ml=(k-1)(k+1)$ we must have $k=6x\pm 1$. Then $ml=x(3x\pm 1).$

Let $m=3^aM$, where $M>1$ is coprime to $3$. Let $x=3^aX$ and then $$Ml=X(3^{a+1}X\pm 1).$$

We can choose $X$, $0< X <M$, such that $3^{a+1}X\equiv 1$ modulo $M$. Replacing $X$ by $M-X$ if necessary we can therefore choose $X$, $0< X \le \frac{M}{2},$ such that $(3^{a+1}X\pm 1)=MY$ for some integer $Y$. Then $l=XY$.

If $m<l$ then $$3^aM<\frac{M}{2}\frac{(3^{a+1}M\pm 2)}{2M} \text{ i.e. } 3^a<\frac{3^{a+1}\pm \frac{2}{M}}{4}. $$ Then $a=0$ and $M=2$. But then $l=1$ gives the solution $12ml+1=5^2$.

If

Let $m=3^a$ and $x=3^aX$, then we must solve $l=X(3^{a+1}X\pm 1)>3^a=m,$ a contradiction.