Six consecutive positive integers with certain shape

Question

Are there 6 consecutive positive integers, where each of them is a square or the product of a prime and a square ?

If they exist, one of those six integers A will be the product of 2 and a square of an odd integer，and another B will be the product of 3 and a square with the form 3k+1. And the difference between A and B may be -1 or 2.

These may cause that there will be only finitely many solutions to the question，or no solution.

However, I don't know how to solve the equations, which requires the six integers equal to a square or a product of a prime and a square.

In fact, except (1,2,3,4,5), (97,98,99,100,101) and (241,242,243,244,245), maybe even 5 consecutive these kind of integers don't exist any more.

8 consecutive these kind of integers do not exist，because there will be two integers among them, both of which equal a product of 2 and a square, and the difference between them is 4. That's impossible for integers.

Motivation: I want to find some consecutive integers in A258456, the product of whose divisors is not a square.

Numbers in this sequence must be a square or a product of a prime and a square.

Here are some examples of 4 consecutive integers in the sequence. 17,18,19,20 47,48,49,50 575,576,577,578 1249,1250,1251,1252 4049,4050,4051,4052 4799,4800,4801,4802 17297,17298,17299,17300 120049,120050,120051,120052 206081,206082,206083,206084 281249,281250,281251,281252 388961,388962,388963,388964 470447,470448,470449,470450 538721,538722,538723,538724

It seems like there are infinitely many instances of 4 consecutive integers in the sequence.

Answer 1

I'd like to add something from the viewpoint of heuristics. The most restrictive condition in this problem seems to be "occurence of a number of $p$-adic valuation $1$ for a small prime $p$" (since this forces the respective number to be of the form $pX^2$). The crucial point, as observed in the other answer, is that with six consecutive numbers, there is necessarily such a condition at the prime $2$ as well as at the prime $3$, leading to a Pell equation, namely either $2k^2+1=3m^2$ or $k^2-1=6m^2$ (with $k$ odd and $n:=2k^2$ being one of the middle two of the six numbers). The point is that the solutions to these Pell equations are of logarithmic density (i.e., the $d$-th solution is roughly of size $10^d$). Now we might try to fill the remaining four numbers of the $6$-tuple by making them ``as prime as possible" (e.g., $n-2$ and $n+2$ have to be divisible by $4$, so we might hope to make $\frac{n-2}{4}$ and $\frac{n+2}{16}$ both prime, or similar). The thing is that the probability for a number of the size roughly $e^d$ to be prime is about $\frac{1}{d}$ by the prime number theorem, so to make all four numbers prime (up to dividing out some necessary small factors coming from divisibility conditions at $2$, $3$ and $5$) should be around $\frac{1}{d^4}$, and so the probability that this would ever happen in some large range of numbers would be approximately some tail of the series $\sum_{d\in \mathbb{N}}\frac{1}{d^4}$, which becomes very small very quickly.

Of course I oversimplified in many ways, first by assuming that certain probabilities are independent (but that is the usual expectation for things like the twin prime conjecture) and then by making the "as prime as possible" assumption (this was laziness on my part; by allowing "prime times arbitrary square" instead of just "prime times necessary very small square", one would get a somewhat denser, but probably not decisively denser set). Still, this may serve as some vague evidence why it becomes quite unlikely that any such $6$-tuples exist, once a certain range has been checked.

EDIT: This first part seems to now have been rendered obsolete by the answers of Max Alekseyev and Tong Lingling, reducing the whole question to finding integral points on a genus-1 curve $2(\epsilon x^2\pm 1)^2+1=3y^2$($\epsilon\in \{1,2\}$)

Also, note the crucial difference when changing from $6$-tuples to $5$-tuples! Then, the local requirement at $3$ vanishes, since one may pick the middle one of the five numbers to be a multiple of $9$. This gets rid of the "density loss" due to Pell, and one may (among certain other possibilities) look for tuples of the form \begin{equation} n=p_1, n+1=50\cdot (2k+1)^2, n+2 = 9p_2, n+3=4p_3, n+4=p_4 \end{equation}, where all the $p_i$ are prime! There are no local obstructions to this system, so given that for a number of asymptotic size (constant times)$k^2$, the probability to be prime is $\frac{1}{2\log(k)}$, the simultaneous occurrence should be expected with probability around (constant times)$\frac{1}{\log(k)^4}$. Summing over all values of $k$, this diverges, which means that heuristically one should expect infinitely many solutions! and indeed, checking up to a range of $n\approx 1.8\cdot 10^{15}$, there are no less than $74$ solutions to this (even somewhat more restricted than asked in the question) version.

EDIT: For $n$ up to $2\cdot 10^{18}$, there are as many as 1172 solutions.

Answer 2

Given your question, it seems that one should study the pair of equations

$\displaystyle 2x^2 - 3y^2 = -1 \quad \text{and} \quad 2x^2 - 3y^2 = 2.$

(In fact, the quadratic field $\mathbb{Q}(\sqrt{6})$ has class number 1, so in fact $2x^2 - 3y^2$ is equivalent to the principal form $x^2 - 6y^2$.)

We have the basic solutions $(1,1)$ and $(1,0)$ respectively for the two equations. Recall we have the Gauss-Dirichlet composition law:

$\displaystyle (2a^2 - 3b^2)(u^2 - 6v^2) = 2a^2 u^2 - 12a^2 v^2 - 3b^2 u^2 + 18b^2 v^2 = 2(au + 3bv)^2 - 3(2av + bu)^2,$

hence all other solutions arise from the fundamental unit: in this case we have $u^2 - 6v^2 = 1$ has the fundamental solution $(u,v) = (5,2)$ corresponding to the algebraic integer $5 + 2 \sqrt{6}$.

Our basic solutions correspond to $(a,b) = (1,1)$ and $(a,b) = (1,0)$ respectively, which gives us two parametrized families of solutions:

$\displaystyle (U,V) = (au + 3bv, 2av + bu) = (u + 3v, u + 2v)$

and

$\displaystyle (U,V) = (u, 2v)$

where $u + v \sqrt{6}$ is a unit in $\mathcal{O}_{\mathbb{Q}(\sqrt{6})}$. By Dirichlet's unit theorem, all of the units are generated by the fundamental unit, and thus take the form

$\displaystyle u_n + v_n \sqrt{6} = (5 + 2 \sqrt{6})^n = \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j} 5^{n - 2j} 6^j + \sqrt{6} \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n}{2j+1} 5^{n-2j-1} 6^j.$

Doing a mod $5$ analysis, we conclude that one of the integers in your list must either be a perfect square divisible by $5$ or $5k^2$, with $k$ co-prime to $5$. This requires us to consider an equation of the form

$\displaystyle |s^2 - 3t^2| = c, |c| \leq 6$

or

$\displaystyle |5s^2 - 3t^2| = c, |c| \leq 6.$

In both cases, we can parametrize the solutions explicitly as above and this gives us an explicit exponential diophantine equation. Such an equation can likely be approached using linear forms in logarithms, which would rule out large solutions. Then potentially one can check the remaining range to either find a non-trivial example, or show that none exist.

Answer 3

This is just to show that in sextuples of interest $2x^2 - 3y^2 = -1$, while difference $2$ is not possible.

First note that neither $|2x^2-2y^2|<6$ nor $|3x^2-3y^2|<6$ is soluble in distinct positive integers $x,y$. It follows that the required sextuple contains exactly one number with $\nu_2 = 1$ and exactly one number with $\nu_3=1$. These numbers are what OP denoted $A$ and $B$. Clearly, these numbers cannot coincide as otherwise they would have a composite core.

Considering septuples modulo $2^2\cdot 3^2=36$, we see that $A=2x^2$ cannot be in $\{6, 10, 22, 30, 34\}$ since $A/2$ must be a square modulo $36/2$, while $B=3y^2$ cannot be in $\{ 6, 15, 21, 24, 30, 33\}$ since $B/3$ must be a square modulo $36/3$. This leaves only two suitable sextuples modulo 36: starting at 0 and at 35, and in both of which $A-B=-1$.

Answer 4

With the analysis from Stanley, Joachim and Max Alekseyev, maybe I can solve my question now. Because two different integers with the form 4k+2 can not appear in these six numbers, when modulo 4, they will congruent to 3,0,1,2,3,0 or 0,1,2,3,0,1. The number A which congruent to 2 modulo 4 must be the product of 2 and a square of an odd number a, so A-2 is always in these six numbers. If A-2 is a product of a prime and a square, using factorization of the difference between the squares of a and 1，it can be showed that there is a square N among a+1, a-1, 2(a+1), 2(a-1) While A+1 is a product of 3 and a square, there will be a Weierstrass equation if N and A+1 are multiplied. There will be finitely many integer solutions or no solution for a.