# Can $\frac{a^4}{b^3}$ be arbitary close to $1$?

I searched for positive integers $a$ and $b$ for which the difference of $a^4$ and $b^3$ is small compared to the powers $a^4$ and $b^3$. The most spectacular example I currently have is

$$9825757^4-2104527924^3=-137318688623$$ so $(a,b)=(9825757,2104527924)$ and $\frac{a^4}{b^3}=1-1.47\cdot 10^{-17}$

Questions :

• Is it true that $a^4-b^3=n$ has finite many integer solutions for every integer $n\ne 0$ ? I am not sure whether Falting's theorem can be applied here.

• Can $\frac{a^4}{b^3}$ be arbitary near to $1$ from above and below ? This seems to be the case, but I have no idea how to prove it.

• How can I find spectacular pairs $(a,b)$ in the above sense efficiently ? I found my example by brute force.

• For any $c$ let $a=c^3$ and $b=c^4$ ... very efficient :) Do you have other conditions on $n$ (and on $a$, $b$) apart from $n\not=0$? – Mirko Apr 23 '17 at 21:35
• @Mirko In your example, we have $n=0$ and $a^4=b^3$, which is of course not allowed. – Peter Apr 23 '17 at 21:41
• If you care, I will update my answer when I find better results – enedil Apr 23 '17 at 23:03
• A few further noteworthy pairs $(a,b)$: $$(a,b)=(27682458, 8374126805); \epsilon = 6.67\cdot10^{-18};$$ $$(a,b)=(66991472, 27207754120); \epsilon = 3.17\cdot10^{-18};$$ $$(a,b)=(85764250, 37821759217); \epsilon = 2.67\cdot10^{-18};$$ $$(a,b)=(113190462,54753654356); \epsilon = 5.62\cdot10^{-19};$$ $$(a,b)=(176874995,99286131022); \epsilon = 1.90\cdot10^{-20};$$ $$(a,b)=(671960665,588561434613); \epsilon = 2.09\cdot10^{-21};$$ $$...$$ – Oleg567 Apr 25 '17 at 5:17
• Empirically, these pairs show that one can expect to find such pairs $(a,b)$ which provide $\epsilon<\dfrac{1}{a^2}$. – Oleg567 Apr 25 '17 at 5:31

Regarding dot number 2 and 3: Find such integers $a$ and $b$ so $a^4=b^3$. It's rather obvious that there are infinitely many of them. Just set $a=c^3$ and $b=c^4$ for some integer $c$. Then the ratio $\frac{(a+1)^4}{b^3}$ is arbitrarily close to $1$. Analogously, $\frac{(a-1)^4}{b^3}$ is close from below. This also gives a method of generating examples.

Edit regarding dot number one:

If $n=1$ I have found a proof that no such solution exist. $$a^4 - b^3 = 1 \Leftrightarrow b^3 = a^4 - 1 = (a^2 + 1)(a^2 - 1) = (a^2 + 1)(a+1)(a-1)$$ I will show that a common divisor of any two of three factors on the right side is at most $2$. Indeed, by Euclid's algorithm $$gcd(a-1, a+1) = gcd(a-1, 2) \in \{ 1, 2\}$$ $$gcd(a-1, a^2+1) = gcd(a-1, a^2 + 1 - 2(a-1)) =$$$$= gcd(a-1, (a-1)^2+2) = gcd(a-1, 2) \in \{ 1, 2\}$$ The third proof is analogous.

Moreover, if $a$ is even, all three of these equal $1$ while for odd $a$, all three equal $2$. If $a$ is odd, then consider $b' = \frac b 2$. Else just set $b' = b$. This yields that integral $b'^3$ is a product of coprime integers, thus all three of them are a cube. However $a+1$ and $a-1$ differ only by $2$, while $\frac{a+1}2$ and $\frac{a-1}2$ differ by $1$, therefore they can't be both cubes, since the least difference between consecutive cubes is $2^3 - 1^3 = 7$.

• Very nice answer (+1)! – Peter Apr 23 '17 at 21:47
• You can also use ${a^4\over (b-1)^3}$ and ${a^4\over (b+1)^3}$ – N74 Apr 23 '17 at 21:50

On question 2.

Taking $$a=k^3-3,\\b=k^4-4k,$$ for $k\in\mathbb{N}$, $k>4$, one can construct sequence of pairs $(a,b)$, for which $$a^4 = b^3+n,$$ where $$n = 6k^6-44k^3+81,$$ so $$n=6a^2-8a+3<6a^2.$$

Therefore, one can construct infinitely many pairs $(a,b)$ which provide error estimation $$\left|\dfrac{b^3}{a^4}-1\right|<\dfrac{6}{a^2}.$$

Other sequence of such kind:

$$a=(3k)^3-1,\\b=(3k)^4-4k,$$ provides better error estimation, but with the same asymptotic (~ $a^{-2}$).

Consider $a,b$ coprime positive numbers. There are $2$ cases:

$$a^4+d=b^3,\tag{1}$$ $$b^3+d=a^4.\tag{2}$$

According to ABC conjecture, there are finitely many pairs $(a,b)$ such that $$d<{a^{5/3-\epsilon}},\tag{3}$$ where $\epsilon>0$.

Explanation.

Denote $C = \max\{a^4,b^3\}$.

$(1)$ and $(2)$ are expressions of the form $$A+B=C.$$

Refer now to to ABC conjecture.

Denote $$R = \mathrm{rad}(ABC),$$ where $\mathrm{rad}(\bullet)$ is radical of an integer.

Then $$R = \mathrm{rad}(abd) \le abd,$$ $$\log R \le \log a + \log b + \log d.$$

Denote $x = C^{1/12}$. Denote $y = \log x = \dfrac{1}{12}\log C$. Then $\log a \approx 3y$, $\log b \approx 4y$.

Consider value $q$ ("quality"):

$$q = \dfrac{\log C}{\log R} \ge \dfrac{\log C}{\log a+ \log b + \log d}\approx \dfrac{12 y}{3y+4y+\log d}.$$

If condition $(3)$ is true, then $\log d < (5-3\epsilon)y$, and $q$ is greater than $1$ (with some gap).

Your example has $\log d \approx 4.7785 y$, which provides quality $1.0188$ for ABC-triple $(9825757^4,137318688623, 2104527924^3)$.

This implies that for any $n\ne 0$ there are finitely many coprime pairs$(a,b)$ s.t. $a^4-b^3=n$. (if ABC conjecture is true, of course).

• You should have stated it beforehand that it might be true that ABC conjecture is false. – enedil Apr 26 '17 at 21:29
• Sure, this fact is included in definition of the word "conjecture". – Oleg567 Apr 27 '17 at 3:14
• that's what I means - it's misleadong a bit – enedil Apr 27 '17 at 5:28

For your first question: My immediate thought was: why wouldn't that be. But I have no idea how to prove it.

As for the ratio $\frac{a^4}{b^3}$, we can obviously make that equal to $1$, by selecting a number $n$ and let $a=n^3$ and $b=n^4$. But if we want it to be strictly larger than $1$ we basically just have to multiply $a$ with $1+\varepsilon$, as that is not an integer, we can "restrict" ourselves to rational $\varepsilon$'s and multiply $n$ with the denominator of $\varepsilon$.

That should be a method for generating example that brings you within a certain distance of $1$.

• The questions $2)$ and $3)$ have already been answered by enedil – Peter Apr 23 '17 at 21:55
• I know. But I was in the process of writing my answer, when he posted his, and my method doesn't just give a way of generating examples, it gives a way of generating examples that bring you within a certain distance of $1$. If you need an example where $\frac{a^4}{b^3} < 1+10^{-474}$, enedil's answer (which I upvoted because it's beautifully simple) doesn't tell you what to do, mine does. – Henrik Apr 23 '17 at 22:06
• Hi Henrik. My solution suggests binary search :P Every time I generate something I think of it as a way to go (as $O(log n)$ is usually not very worse than $O(1)$ – enedil Apr 23 '17 at 22:21