Can we always find values $(a,d)$ such that $\frac{d+an}{4a-1}$ is an integer for all $n$

Question

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.

Answer 1

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 $5$.

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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 $d\mid (an)^2$. $\quad\blacksquare$

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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 : $$\frac{d+an}{4a-1}=\frac{2s(12k+1)}{4s-1}=2s\times (\text{an integer})$$

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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 $4s-1$

Answer 2

I have a partial result.

For $n=3q+r$ with $r\in\{0,1,2\}$

We have for $a=1$ $$F=\dfrac{an+d}{4a-1}=\dfrac{3q+r+d}3=q+\dfrac{d+r}3$$

• For $r=0$ then $d=3\qquad$ [ $n^2=(3q)^2$ so $d\mid n^2$ ]

• For $r=2$ then $d=1\qquad$ [ $d=1$ always divides $n^2$ ]

• For $r=1$ then $d=2$ and $q\text{ odd}\qquad$ [ $n^2\equiv q+1\pmod{2}$, so if $q$ odd $d=2$ divides $n^2$ ]

In all three cases then $d+r=3$ so $F$ is an integer.

The only difficult case appears to be when $n=6p+1$.

For that case, the simulation on computer does not show a good pattern for $a$ or $d$.

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Edit 28/11

Inspired by mathlove post I took my chance at endings $1,9$ in decimal or or $k=5m,5m+2$.

The most promising road was $\begin{cases}a=uv\\d=un\end{cases}$

Resulting in $\dfrac{u(1+v)}{4uv−1}\in\mathbb N$.

Unfortunately the numerator is too small and this doesn't work out.

Then I explorer another idea with computer simulation $\begin{cases}a=uv\\d=uan^2\end{cases}$

In this case $f(a,d,n)=\dfrac{an+d}{4a-1}=an\times\dfrac{1+un}{4uv-1}=an\times g(u,v,n)$.

Still for the $n=24k+1$ primes, this reduced problem $g(u,v,n)\in\mathbb N$ seems almost always to have a solution in $(u,v)$

Only very few cases actually do not work (either impossible or u,v were possibly too large).

$k\in\{8,105,2772,3185,10500,13217,13728,16875,50547,59892,72140,81593,83252,...\}$

I don't know what particularity these numbers share, maybe it is a good place to dig in, or maybe it also means that we have to go for another approach. Anyway, if anyone is tempted to explore further, this is what to expect.

Answer 3

The Erdös-Straus conjecture is that for every $n>1\in\mathbb{N}$, there are $x, y, z\in\mathbb{N}$ where \begin{align} \frac4{n} &= \frac1{x}+\frac1{y}+\frac1{z}.\tag{ES} \end{align} If $n$ is a composite $qr$, then we may consider the case $n=q$ instead, then multiply each denominator by $r$. It is therefore sufficient to consider prime $n$.

In the terminology and notation of Mordell (also given in Elsholz & Tao), in a type-II solution, two of the denominators $x, y$ and $z$ are multiples of $n$. Mordell proves that, if $n$ is prime, such a solution has the form $$\frac4{n} = \frac1{abd}+\frac1{acdn}+\frac1{bcdn},\tag{M}$$ that is, after the denominators are divided by their highest common factor $d$, the product of the resulting numbers is a square $(abcn)^2$.

A stronger variant of the Erdös-Straus conjecture is that every $n$ has a type-II solution. No counterexamples are known.

Greg's conjecture seems to be a weaker variant of the above stronger variant. I replace the OP's $a$ and $d$ with $A$ and $D$ to avoid confusion with Mordell's. The OP's condition on an expression being an integer $u$ means that \begin{align*} u(4A-1) &= An+D\\ \implies 4Au &= An+D+u\\ \implies \frac4{n} &= \frac1{u}+\frac{D}{Aun}+\frac1{An}\tag{G} \end{align*} which is a type-2 EFR of $4/n$ with $x=u, y=Aun/D, z=An$, provided that $D\mid Aun$. Greg generously allows $D\mid A^2 n^2$. However, if $D\nmid Au$, then $n\nmid y$ and this is actually a type-1 EFR.

Equating the respective terms in equations (ES), (M) and (G) implies \begin{align*} u &= x=abd\\ Aun/D &= y=acdn \implies Au=acdD\\ An &= z=bcnd \implies A=bcd\\ \implies acdD &= Au\\ &= (bcd)(abd)\\ \implies D &= b^2d \end{align*} so $D\mid A^2$ and a fortiori $D\mid A^2 n^2$.

C. Elsholtz and T. Tao. Counting the number of solutions to the Erdös–Straus equation on unit fractions. 24 Oct 2018.