Erdős Problem #364

Erdős Problem #364: three consecutive powerful integers

1. Statement and scope

A positive integer is powerful, or $2$-full, when every prime exponent in it is either zero or at least two. Are there any three consecutive positive powerful integers? We prove structural restrictions, reconstruct the infinite family of consecutive pairs, and give the precise conditional consequence of $abc$. An exact finite search is included as a bounded check, not as an answer to the infinite question, which remains unresolved.

2. Unique parametrization and an exhaustive search

Every powerful integer has a unique expression \[ n=a^2b^3,\qquad a,b\ge1,\quad b\text{ squarefree}. \] For each odd positive prime exponent $e$, place that prime once in $b$ and $(e-3)/2$ times in $a$; for an even exponent use $e/2$ copies in $a$. This construction is forced by parity and proves uniqueness. There is no requirement that $a$ and $b$ be coprime. In particular the number of powerful integers at most $X$ is at most \[ \sum_{b\le X^{1/3}}\left\lfloor\sqrt{X/b^3}\right\rfloor \le \sqrt X\sum_{b\ge1}b^{-3/2}. \] Enumerating precisely the squarefree $b\le X^{1/3}$ and $1\le a\le\lfloor\sqrt{X/b^3}\rfloor$ therefore gives an exhaustive finite algorithm, with no factorization heuristic. For $X=10^{10}$ the exact enumeration contains $214122$ integers. Checking membership of $n,n+1,n+2$ finds no triple all of whose members lie in $[1,X]$. This computation makes no claim about larger integers and is far smaller than published computational searches.

3. Local obstructions and an infinite family of pairs

Four consecutive integers cannot all be powerful, because one has residue $2$ modulo $4$. If $m-1,m,m+1$ are powerful, this also shows that the center satisfies $m\equiv0\pmod4$. Residues $3,6$ modulo $9$ are impossible for a powerful integer, so examining the three consecutive residues gives \[ m\equiv0,1,8\pmod9. \] The two odd neighbors cannot both be squares: two positive squares differing by $2$ would give $(v-u)(v+u)=2$, which is impossible for positive integers $u<v$. Thus at least one neighbor has a nontrivial squarefree cube factor in the parametrization above.

These restrictions do not prevent consecutive pairs. Starting at $(x_1,y_1)=(3,1)$, define \[ x_{j+1}=3x_j+8y_j,\qquad y_{j+1}=x_j+3y_j. \] Direct expansion preserves $x_j^2-8y_j^2=1$, and both coordinates increase. Hence $8y_j^2$ and $8y_j^2+1=x_j^2$ give infinitely many distinct powerful pairs. Every prime exponent in the first number is even except possibly that of $2$, which is $3+2v_2(y_j)$ and therefore at least three.

4. What the $abc$ conjecture would actually imply

Assume the usual $abc$ conjecture: for every $\varepsilon>0$ there is $C_\varepsilon$ such that coprime positive $a+b=c$ satisfy $c\le C_\varepsilon\operatorname{rad}(abc)^{1+\varepsilon}$. If $m-1,m,m+1$ are powerful, each obeys $\operatorname{rad}(u)\le\sqrt u$. Apply the conjecture to \[ (m^2-1)+1=m^2. \] The summands are coprime, and \[ \operatorname{rad}\big((m^2-1)m^2\big) =\operatorname{rad}\big((m-1)m(m+1)\big) \le\sqrt{(m-1)m(m+1)}<m^{3/2}. \] It follows that $m^2\le C_\varepsilon m^{3(1+\varepsilon)/2}$. Taking $\varepsilon<1/3$ bounds $m$. Thus $abc$ implies only finitely many such triples, not that there are none. No use of this conjecture is made in the unconditional results or the finite search.

5. Assessment and references

The remaining obligation is an unconditional existence proof or an obstruction valid for every candidate center. The congruences, parametrization, and conditional finiteness do not provide it. Source: https://www.erdosproblems.com/364, checked 2026-09-25, which records Mahler's Pell observation and the conditional $abc$ consequence. The prior GPT Pro 5.2 argument was checked; its parametrization and elementary obstruction are reconstructed here. No novelty or formal Lean verification is claimed.

COMPLETION ESTIMATE

COMPLETION: 35%

1) FORMAL RESTATEMENT

A positive integer $n$ is powerful (a.k.a. 2-full, squareful) if \[ \forall\text{ primes }p:\ p\mid n\ \Rightarrow\ p^2\mid n. \]

The problem asks:

Does there exist $N\in\mathbb N$ such that $N,\,N+1,\,N+2$ are all powerful?

(Equivalently: do there exist three consecutive positive integers all powerful?)

Edge cases: * $1$ is powerful by the given definition (no prime divides 1). * $2$ is not powerful.

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2) QUICK LITERATURE/CONTEXT CHECK

I only record what the provided problem text states (no additional external claims):

* There are infinitely many pairs of consecutive powerful numbers from solutions to $x^2=8y^2+1$. * The abc conjecture would imply only finitely many such triples. * There are no quadruples of consecutive powerful numbers. * A computational claim (OEIS A060355) is mentioned: no such triples with $n<10^{22}$.

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3) ATTACK PLAN

Proof track (partial structure). 1. Prove standard structural characterizations of powerful numbers. 2. Prove rigorously the infinite consecutive-pairs construction from Pell’s equation. 3. Prove the easy obstruction for 4 consecutive powerful numbers.

Disproof track. Try to find a triple by computation in a modest range. (No triple found up to $10^6$.)

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4) WORK

Phase 1: FAST REALITY CHECK (computation)

I brute-searched $N\le 10^6$ and found no triple $(N,N+1,N+2)$ with all three powerful.

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Lemma 4.1 (Powerful $\Leftrightarrow a^2 b^3$ with $b$ squarefree)

A positive integer $n$ is powerful if and only if it can be written as \[ n=a^2 b^3 \] with $a,b\in\mathbb N$ and $b$ squarefree.

Proof.

($\Rightarrow$). Write the prime factorization $n=\prod_p p^{e_p}$ with each $e_p\ge 0$. Powerfulness means: whenever $e_p>0$, we have $e_p\ge 2$.

Define \[ b:=\prod_{p: e_p\text{ odd}} p\quad\text{(so }b\text{ is squarefree),}\qquad a:=\prod_p p^{(e_p-3\cdot\mathbf 1_{e_p\text{ odd}})/2}. \] This makes sense because: * if $e_p$ is even, the exponent is $e_p/2\in\mathbb N_0$; * if $e_p$ is odd and $e_p\ge 3$, then $(e_p-3)/2\in\mathbb N_0$.

Then for each prime $p$, the exponent of $p$ in $a^2 b^3$ is $2\cdot\frac{e_p-3\cdot\mathbf 1_{e_p\text{ odd}}}{2} + 3\cdot\mathbf 1_{e_p\text{ odd}}=e_p$. Hence $n=a^2 b^3$.

($\Leftarrow$). If $n=a^2 b^3$, then for any prime $p\mid n$, either $p\mid a$ (then $p^2\mid a^2\mid n$) or $p\mid b$ (then $p^3\mid b^3\mid n$, so in particular $p^2\mid n$). Thus $n$ is powerful.

This proves the equivalence. $\square$

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Lemma 4.2 (Infinite consecutive pairs from Pell)

Let $(x,y)\in\mathbb N^2$ satisfy the Pell equation \[ x^2-8y^2=1. \] Then $8y^2$ and $x^2$ are consecutive powerful numbers.

Proof. The Pell equation gives $x^2=8y^2+1$, so $x^2$ and $8y^2$ differ by 1.

* $x^2$ is a perfect square, hence for every prime $p\mid x^2$ we have $p^2\mid x^2$; thus $x^2$ is powerful. * $8y^2 = 2^3\cdot y^2$. Any prime divisor of $y$ occurs in $y^2$ with exponent at least 2, and the prime 2 occurs with exponent 3. Hence every prime divisor of $8y^2$ has exponent at least 2, so $8y^2$ is powerful.

To see there are infinitely many positive integer solutions, start from the solution $(x_1,y_1)=(3,1)$ (since $3^2-8\cdot 1^2=1$) and define recursively \[ x_{t+1}=3x_t+8y_t,\qquad y_{t+1}=x_t+3y_t. \] If $x_t^2-8y_t^2=1$, then \[ x_{t+1}^2-8y_{t+1}^2=(3x_t+8y_t)^2-8(x_t+3y_t)^2=x_t^2-8y_t^2=1, \] so $(x_{t+1},y_{t+1})$ is also a solution. Moreover $x_{t+1}=3x_t+8y_t>x_t$, hence the solutions are all distinct and infinite in number.

Therefore there are infinitely many such consecutive powerful pairs. $\square$

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Lemma 4.3 (No four consecutive powerful numbers)

There do not exist four consecutive positive integers all of which are powerful.

Proof. Among any four consecutive integers, exactly one is congruent to $2\pmod 4$. Such an integer has 2-adic valuation exactly 1, i.e. it is divisible by 2 but not by 4. Hence for that integer $n$, the prime $2\mid n$ but $2^2\nmid n$, so $n$ is not powerful. $\square$

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5) VERIFICATION

* Lemma 4.1: the construction of $a,b$ assigns odd exponents to $b$ (once each) and the remaining even part to $a^2$; the exponent check is explicit prime-by-prime. * Lemma 4.2: the only property used is “square implies powerful” and “$y^2$ gives exponent $\ge 2$” plus the explicit exponent 3 of 2 in 8. * Lemma 4.3: the mod-4 argument is airtight. * Computation is only a sanity check; no conclusions beyond $10^6$ are claimed.

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6) FINAL

UNRESOLVED

(i) Strongest fully proved partial result obtained here.

* There are infinitely many pairs of consecutive powerful numbers of the form $(8y^2, x^2)$ coming from Pell solutions $x^2-8y^2=1$ (Lemma 4.2). * There are no four consecutive powerful numbers (Lemma 4.3). * A brute-force search finds no triple $(N,N+1,N+2)$ of powerful numbers with $N\le 10^6$.

(ii) Exact first gap.

Prove or disprove: \[ \exists N\in\mathbb N:\ N,\ N+1,\ N+2\text{ are all powerful}. \]

(iii) Top 3 next moves (concrete targets).

1. Use Lemma 4.1 to parametrize a hypothetical triple as $N=a^2b^3$, $N+1=c^2d^3$, $N+2=e^2f^3$ with squarefree $b,d,f$, and exploit gcd constraints $\gcd(N,N+1)=\gcd(N+1,N+2)=1$. 2. Search for strong modular obstructions: for a powerful $n$, $n\equiv 0,1\pmod 4$; combine such restrictions across three consecutive integers to narrow cases. 3. If allowed to assume abc (as the problem text mentions), formalize the implication “finitely many triples” and try to obtain an unconditional weaker bound (e.g. growth gaps between powerful numbers).

(iv) What a minimal counterexample would likely look like.

If a triple exists, the smallest $N$ must be very large (computational evidence suggests no small examples). Structurally, by coprimality, the three powerful numbers would have disjoint prime sets pairwise, forcing a delicate arrangement where each of $N, N+1, N+2$ is composed entirely of prime powers $\ge 2$ yet adjacent numbers share no primes.