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

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

To find them is simple. $quad knotinmathbb{N} text { for }k=sqrt{24 m + 1)}$.

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

This is the same as testing $quadsqrt{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 $quad 24mtimes(l)+1, lle m.quad$ For examples,

begin{align*} text{If }l=2,quad& min{ emptyset }\ text{If }l=3,quad& min{ 1,2,3}\ text{If }lin{4,5,6,7,8}quad& min{ emptyset }\ text{If }l=9,quad& min{ 1,2,3,4,5,6,7,8,9}\ text{If }lin{10,11,12,cdots, 23}quad& min{ 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.

$$l = frac{k^2 - 1}{24 m} quadtext{and we know }mne0tag{2}$$

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.