TITLE: Can we always find values $(a,d)$ such that $\frac{d+an}{4a-1}$ is an integer for all $n$
QUESTION [8 upvotes]: I'm working on determining if a certain form of simple hyperbolic diophantine equation always has solutions. I did some work and determined the answer is yes if the following fraction can always be an integer for some $a$ and some $d$ dividing $(an)^2$.

For all $n$, can we always find values $(a,d)$ such that $\dfrac{d+an}{4a-1}$ is an integer and where, 

$n$ is any fixed integer $\geq2$
$a$ is any positive integer
$d$ is a positive divisor of $(an)^2$.

Required: One need to find at least one couple $(a,d)$ per value of $n$ verifying the conditions.

I have checked for values of $n$ up to one million using a computer. Does anybody have any ideas how to show this is true or have a counterexample? 
By the way, this purely recreational. Thanks for any help or ideas!
Edit: Sorry for the confusion. I only care that some $a$ and some $d$ dividing $(an)^2$ exist such that the fraction becomes an integer. But $a$ can take on any positive integer value.
Update: I have made some progress with the problem. I can show the statement holds for all integers not of the form $n=24k+1$. Furthermore, I can show that if the statement holds for a composite integer's prime factors, it holds for the composite number itself. So, I have narrowed the problem down to primes $n=p=24k+1$. Hopefully somebody has an idea to deal with the primes.

REPLY [5 votes]: This is a partial answer. 
This answer deals with the case for $n=6k+1$ since zwim has already dealt with the cases for $n=6k+r$ where $r$ is either $0,2,3,4$ or $5$.

I think some results you've written in "Update" are important, so let me add my proof for them here. 

I can show the statement holds for all integers not of the form $n=24k+1$. 

Proof : 
If $n=24k+7$, then since $n\equiv -1\pmod 4$, there is at least one prime $p$ of the form $4s-1$ such that $p\mid n$, so $\frac{d+an}{4a-1}$ is an integer for $(a,d)=(s,n)$ as follows :$$\frac{d+an}{4a-1}=\frac{n+sn}{4s-1}=\frac{n(s+1)}{4s-1}=(s+1)\times (\text{an integer})$$
If $n=24k+13$, then since $n+1=2(12k+7)$ with $12k+7\equiv -1\pmod 4$, there is at least one prime $p$ of the form $4s-1$ such that $p\mid 12k+7$, so $\frac{d+an}{4a-1}$ is an integer for $(a,d)=(s,sn^2)$ as follows : $$\frac{d+an}{4a-1}=\frac{sn^2+sn}{4s-1}=\frac{sn(n+1)}{4s-1}=2sn\times (\text{an integer})$$ 
If $n=24k+19$, then we can find $(a,d)$ in the similar way as the case where $n=24k+7$. $\quad\blacksquare$

I can show that if the statement holds for a composite integer's prime factors, it holds for the composite number itself.

Proof : Let $n=pq$. If there is a pair of positive integers $(a_1,d_1)$ such that $\frac{d_1+a_1p}{4a_1-1}$ is an integer with $d_1\mid (a_1p)^2$, then $\frac{d+an}{4a-1}$ is an integer for $(a,d)=(a_1,d_1q)$ satisfying $d\mid (an)^2$. $\quad\blacksquare$ 

Next, let us show some partial results for primes $n=24k+1$. (Michael's comments are helpful here.)

If $n=24k+1$ whose right-most digit is $3$ in decimal representation ($n$ is not necessarily prime), then solving $24k+1\equiv 3\pmod{10}$, we see that there is an integer $m$ such that $k=5m+3$, and that $\frac{d+an}{4a-1}$ is an integer for $(a,d)=(4,8)$ as follows : $$\frac{d+an}{4a-1}=\frac{8+4(24(5m+3)+1)}{4\cdot 4-1}=32m+20$$
If $n=24k+1$ whose right-most digit is $7$ in decimal representation ($n$ is not necessarily prime), then solving $24k+1\equiv 7\pmod{10}$, we see that that there is an integer $m$ such that $k=5m+4$, and that $\frac{d+an}{4a-1}$ is an integer for $(a,d)=(4,2)$ as follows : $$\frac{d+an}{4a-1}=\frac{2+4(24(5m+4)+1)}{4\cdot 4-1}=32m+26$$
If $n=24k+1$ is prime, then we may suppose that $d$ is either $1,a,a^2,a',a'n$ or $a'n^2$ where $a'\ (\not=1,a,a^2)$ is a divisor of $a^2$ because of the followings where we use that $\gcd(2an,4a-1)=1$ :
$$\begin{align}&\text{If $d=1$ or $d=a^2n^2$, then $\frac{d+an}{4a-1}\in\mathbb Z$ if and only if $\frac{an+1}{4a-1}\in\mathbb Z$}\\\\&
\text{If $d=a$ or $d=an^2$, then $\frac{d+an}{4a-1}\in\mathbb Z$ if and only if $\frac{12k+1}{4a-1}\in\mathbb Z$}\\\\&
\text{If $d=a^2$ or $d=n^2$, then $\frac{d+an}{4a-1}\in\mathbb Z$ if and only if $\frac{a+n}{4a-1}\in\mathbb Z$}\\\\&\text{If $d=n$ or $d=a^2n$, then $\frac{d+an}{4a-1}\in\mathbb Z$ if and only if $\frac{a+1}{4a-1}\in\mathbb Z$ which cannot be an integer}\\\\&
\text{If $d=an$, then $\frac{d+an}{4a-1}=\frac{2an}{4a-1}$ cannot be an integer}\end{align}$$
If $n=24k+1$ is prime and $12k+1$ has at least one prime factor of the form $4s-1$, then $\frac{d+an}{4a-1}$ is an integer for $(a,d)=(s,s)$ as follows : $$\frac{d+an}{4a-1}=\frac{2s(12k+1)}{4s-1}=2s\times (\text{an integer})$$


So, the remaining cases are primes of the form $n=24k+1$ such that

The right-most digit in decimal representation is either $1$ or $9$
$12k+1$ has no prime factors of the form $4s-1$