Erdős Problem #369

FORMAL RESTATEMENT

Erdős #369; reconstruction of the smooth-polynomial construction, 2026-09-24. For fixed $\varepsilon>0$ and $k\geq2$, show that every sufficiently large $N$ contains $k$ consecutive positive integers all of whose prime factors are at most $N^\varepsilon$.

QUICK LITERATURE AND CONTEXT CHECK

Taken literally, the question allows the fixed block $1,\ldots,k$. The substantive stronger statement is that such a block can be placed near $N$. Balog--Wooley established arbitrarily large smooth blocks; Theorem 2.1 of Bober--Fretwell--Martin--Wooley, Smooth values of polynomials, supplies a polynomial construction. The argument below reconstructs its linear-factor case and obtains a block inside $[N-N^c,N]$ for some fixed $c<1$. This is a reconstruction of known mathematics.

OBJECTIVE AND ATTACK PLAN

Choose disjoint prime sets so that several cyclotomic factorizations simultaneously have small degree. Evaluate the resulting polynomial at its last integer argument below $N$.

WORK

Choosing the exponents. Put $\eta=\min(\varepsilon/2,1/2)$. The divergence of the sum of reciprocals of the primes implies that there are disjoint finite sets $\mathcal P_1,\ldots,\mathcal P_k$ of odd primes such that, on writing \[ \gamma_j=\prod_{p\in\mathcal P_j}p,\quad D=\prod_{j=1}^k\gamma_j,\quad D_j=D/\gamma_j, \] one has $\varphi(\gamma_j)/\gamma_j<\eta$ for every $j$. Indeed, $\prod(1-1/p)\leq\exp(-\sum1/p)$, and deleting finitely many primes leaves the reciprocal sum divergent. All $\gamma_j$ are pairwise coprime. Choose $1\leq u_j<\gamma_j$ satisfying $D_j u_j\equiv1\pmod {\gamma_j}$, and put $\mu_j=D_j u_j$. Thus $\mu_j\equiv1\pmod {\gamma_j}$ and $\gamma_i\mid\mu_j$ if $i\ne j$. Define \[ C=\prod_{j=1}^k j^{\mu_j},\qquad g(t)=Ct^D. \] For each $j$ the positive integer \[ A_j=j^{(\mu_j-1)/\gamma_j}\prod_{i\ne j}i^{\mu_i/\gamma_j} \] is well defined and satisfies $C=jA_j^{\gamma_j}$. Consequently \[ g(t)-j=j\big((A_jt^{D_j})^{\gamma_j}-1\big) =j\prod_{e\mid\gamma_j}\Phi_e(A_jt^{D_j}), \tag{1} \] where $\Phi_e$ denotes the $e$th cyclotomic polynomial.

Bounding every prime factor. Because $\gamma_j$ is squarefree, $e\mid\gamma_j$ implies $\varphi(e)\leq\varphi(\gamma_j)$. Each polynomial factor on the right of $1$ therefore has degree at most \[ D_j\varphi(\gamma_j)<\eta D. \] There are finitely many fixed factors. For all sufficiently large integer $t$, every such factor is nonzero and has absolute value at most $K t^{\eta D}$ for one fixed $K$. Every prime dividing $g(t)-j$ divides either $j$ or one of those integer factors. Since $g(t)=Ct^D$, and $\eta<\varepsilon$, \[ P(g(t)-j)\leq\max(k,Kt^{\eta D})\leq g(t)^\varepsilon \quad (1\leq j\leq k) . \tag{2} \] No assertion about irreducibility after substitution is needed: an integer's prime divisors are bounded by its absolute value.

Positioning the block for every large $N$. Take $t=\lfloor(N/C)^{1/D}\rfloor$. Then $g(t)\leq N<g(t+1)$, and \[ 0\leq N-g(t)<C\big((t+1)^D-t^D\big)=O(N^{1-1/D}). \] The $k$ consecutive numbers $g(t)-k,\ldots,g(t)-1$ are positive for large $N$ and satisfy $2$, hence are $N^\varepsilon$-smooth. With $c=1-1/(2D)$, their distance from $N$ is eventually at most $N^c$. This proves the stronger all-$N$ statement.

RESULT AND LIMITATIONS

Attempt status: solved. The construction and the passage from polynomial values to a block near every large $N$ are proved above. The standard facts used are divergence of the prime reciprocal sum and cyclotomic factorization. No effective small threshold or optimal value of $c$ is claimed.

SOURCES

https://www.erdosproblems.com/369; Bober, Fretwell, Martin and Wooley, Smooth values of polynomials, Theorem 2.1: https://arxiv.org/html/1710.01970. Tao's explanation of the near-$N$ deduction: https://www.erdosproblems.com/forum/discuss/369.

Completion rate estimate: 100%.

Erdős Problem 369

Problem statement.

Let $\varepsilon>0$ and $k\geq 2$. Is it true that, for all sufficiently large $n$, there is a sequence of $k$ consecutive integers in $\{1,\ldots,n\}$ all of which are $n^{\varepsilon}$-smooth?

Tools in this round.

Web browsing: YES. Computation: YES.

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. Round 1 already gave a complete proof of the literal statement, so the only mathematically meaningful Round-2 task is to identify the exact defect in the formulation and determine which corrected form actually captures a nontrivial asymptotic problem.

The plan is to separate two possible repairs:

• replace "$n^{\varepsilon}$-smooth" by "$m^{\varepsilon}$-smooth" for each term $m$ of the block;
• force the block itself to lie near the top of the interval, for example inside $[n/2,n]$.

I will prove that the first repair is true (indeed much weaker than it first appears), whereas the second repair is the one that preserves the intended asymptotic content and remains open.

2) Round-1 FOUNDATION USED

I use the following Round-1 results.

• Round 1 proved the literal statement exactly as written by taking the fixed block $1,2,\ldots,k$ once $n\geq \max\{k,k^{1/\varepsilon}\}$.
• Round 1 identified the core ambiguity: the phrase "$n^{\varepsilon}$-smooth" ties the smoothness threshold to the ambient parameter $n$, not to the size or location of the block itself.
• Round 1 also noted two natural repairs already discussed in the literature: termwise $m^{\varepsilon}$-smoothness, and a localization condition such as placing the block in $[n/2,n]$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new point in this round is the following adversarial observation.

Merely replacing $n^{\varepsilon}$-smooth by $m^{\varepsilon}$-smooth does not restore genuine asymptotic dependence on $n$. Once a single admissible block exists anywhere, the same block lies in $\{1,\ldots,n\}$ for every larger $n$. So this repaired statement is equivalent simply to the existence of one admissible block.

This means that the true logical defect in the original statement is not only the global smoothness threshold; it is also the absence of any location condition. Balog--Wooley's theorem then settles the "termwise-smoothness" repair immediately, while the localization repair remains open.

4) ATTACK PLAN (ROUND-2)

The precise gaps left after Round 1 are:

• the corrected variants were only identified informally, not resolved;
• it was not yet shown which corrected variant preserves the asymptotic flavor of the original question.

I will now prove three claims.

1. The termwise correction \[ (\forall X \text{ large})\; \exists \text{ a block of $k$ consecutive integers in } \{1,\ldots,X\} \] whose members $m$ all satisfy $P^{+}(m)\leq m^{\varepsilon}$ is equivalent to the existence of at least one such block.
2. This corrected statement is true, by Balog--Wooley.
3. Therefore the only repair that keeps the parameter $n$ genuinely relevant is a localization requirement such as $P\subseteq [n/2,n]$; that version remains open.

5) WORK (ROUND-2)

Write $P^{+}(m)$ for the largest prime factor of $m$.

Step 1: the smoothness-only repair collapses to a one-time existence statement.

Consider the statement \[ \mathcal C_{\mathrm{smooth}}(\varepsilon,k):\qquad \text{for all sufficiently large } X, \text{ there exists a block of } k \text{ consecutive integers in } \{1,\ldots,X\} \] all of whose members $m$ satisfy $P^{+}(m)\leq m^{\varepsilon}$.

I claim that $\mathcal C_{\mathrm{smooth}}(\varepsilon,k)$ is equivalent to the simpler assertion: \[ \mathcal E(\varepsilon,k):\qquad \text{there exists at least one block of } k \text{ consecutive positive integers } \] all of whose members $m$ satisfy $P^{+}(m)\leq m^{\varepsilon}$.

Indeed, if $\mathcal C_{\mathrm{smooth}}(\varepsilon,k)$ holds then certainly $\mathcal E(\varepsilon,k)$ holds. Conversely, suppose one block \[ M+1,M+2,\ldots,M+k \] satisfies $P^{+}(M+j)\leq (M+j)^{\varepsilon}$ for every $1\leq j\leq k$. Then the same block is contained in $\{1,\ldots,X\}$ for every $X\geq M+k$. Hence $\mathcal C_{\mathrm{smooth}}(\varepsilon,k)$ holds.

So the "for all sufficiently large $X$" quantifier is not the real issue here: without a location constraint, one admissible block is enough forever.

Step 2: prove $\mathcal E(\varepsilon,k)$.

There are two cases.

Case 1: $\varepsilon\geq 1$. For every positive integer $m$ we have $P^{+}(m)\leq m\leq m^{\varepsilon}$. Therefore every block of consecutive integers works; in particular $1,2,\ldots,k$ is admissible.

Case 2: $0<\varepsilon<1$. Put $u=1/\varepsilon$, so $u>1$. Balog--Wooley prove that if \[ t(N)=\left\lfloor \frac{\log_{4} N}{\log(3u)}\right\rfloor, \] then for infinitely many positive integers $N$, none of the prime factors of any of the integers \[ N+1,N+2,\ldots,N+t(N) \] exceeds $N^{1/u}=N^{\varepsilon}$.

Since $t(N)\to\infty$, there exists such an $N$ with $t(N)\geq k$. Fix one. Then for every $1\leq j\leq k$, \[ P^{+}(N+j)\leq N^{\varepsilon}< (N+j)^{\varepsilon}, \] because $\varepsilon>0$ and $N+j>N$. Hence the block \[ N+1,N+2,\ldots,N+k \] satisfies $P^{+}(m)\leq m^{\varepsilon}$ termwise. Thus $\mathcal E(\varepsilon,k)$ holds.

By Step 1, $\mathcal C_{\mathrm{smooth}}(\varepsilon,k)$ holds as well.

Step 3: identify the genuinely asymptotic correction.

Now consider instead the localized variant \[ \mathcal C_{\mathrm{loc}}(\varepsilon,k):\qquad \text{for all sufficiently large } X,\text{ there exists a block of } k \text{ consecutive integers } \] contained in $[X/2,X]$ all of which are $X^{\varepsilon}$-smooth.

The monotonicity trick from Step 1 no longer applies: an early block does not stay inside $[X/2,X]$ as $X\to\infty$. Thus the parameter $X$ again becomes essential. This is therefore the correction that preserves the intended asymptotic content. According to the current literature summary checked in this round, this localized version is open for "all sufficiently large $X$".

6) ADVERSARIAL VERIFICATION

I check the vulnerable points explicitly.

• Did the termwise repair really collapse to one-time existence? Yes. The repaired statement asks only for a block somewhere in the initial segment $\{1,\ldots,X\}$. Once one admissible block exists, every larger initial segment still contains it.
• Did I use monotonicity correctly when $0<\varepsilon<1$? Yes. The function $x\mapsto x^{\varepsilon}$ is increasing for every $\varepsilon>0$, so $N^{\varepsilon}<(N+j)^{\varepsilon}$.
• Does Balog--Wooley give length exactly $k$? Not exactly, but it gives length $t(N)$ with $t(N)\to\infty$, so for fixed $k$ there are admissible strings of length at least $k$, and the first $k$ terms give the desired block.
• Is the localization correction really immune to the Round-1 trivialization? Yes. A block near the origin does not remain in $[X/2,X]$ once $X$ grows, so the quantifier "for all sufficiently large $X$" regains genuine content.

No contradiction with Round 1 appears; the new work sharpens the logical diagnosis rather than undoing any previous lemma.

7) FINAL

FULL SOLUTION -- FULL PROOF.

The statement as written is fatally underconstrained and is trivially true, exactly as Round 1 showed. In Round 2 I proved more:

• the smoothness-only repair (replace $n^{\varepsilon}$-smooth by $m^{\varepsilon}$-smooth) is also true, because it is equivalent to the existence of one admissible block and Balog--Wooley provide such blocks;
• the asymptotically meaningful repair is the localized one (for example, requiring the block to lie in $[n/2,n]$), and that is the version that remains open.

Thus the original problem cannot be the intended hard problem in its literal form.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 100%

9) REFERENCES

• A. Balog and T. D. Wooley, On strings of consecutive integers with no large prime factors, J. Austral. Math. Soc. Ser. A 64 $1998$, 266--276.
• T. F. Bloom, Erdős Problem #369, accessed 2026-03-07.

10) OUTPUT FORMAT

Erdős problem number: $x=369$. This section is written as a drop-in LaTeX fragment and is included in the combined file 369_371_373_round2_solutions.tex.

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Problem 369

1) FORMAL RESTATEMENT

Fix real $\varepsilon>0$ and an integer $k\ge 2$. For $n\in\mathbb{N}$, call an integer $m\ge 1$ $n^{\varepsilon}$-smooth if every prime divisor $p\mid m$ satisfies $p\le n^{\varepsilon}$ (equivalently $P(m)\le n^{\varepsilon}$, where $P(m)$ denotes the largest prime factor of $m$ and we set $P(1)=1$).

Claim (literal statement). There exists $n_0(\varepsilon,k)$ such that for every $n\ge n_0(\varepsilon,k)$ there is a block of $k$ consecutive integers contained in $\{1,2,\dots,n\}$, all of which are $n^{\varepsilon}$-smooth.

Misstatement/ambiguity note. The literal claim is elementary because the block may be chosen near $1$ and thus does not "scale with $n$." A minimal nontrivial correction (consistent with how such questions are usually posed) is for example:

Corrected variant (one natural choice). Fix $\varepsilon>0$ and $k\ge 2$. Is it true that for all sufficiently large integers $m$ there exist $k$ consecutive integers $m,m+1,\dots,m+k-1$ such that each is $m^{\varepsilon}$-smooth (i.e. $P(m+j)\le m^{\varepsilon}$ for $0\le j\le k-1$)?

2) KEY DEFINITIONS AND KNOWN FACTS

• $m$ is $y$-smooth means $P(m)\le y$.
• If $1\le m\le y$, then $m$ is $y$-smooth because every prime factor of $m$ is $\le m\le y$.

Quick literature/context check (browsing available).

The corrected (nontrivial) variant above is known to hold for infinitely many starting points $m$ for every fixed $k$ and every $\varepsilon>0$ (Balog--Wooley). Whether such blocks exist for all sufficiently large $m$ (and stronger "near-$n$" variants, e.g. constrained to $[n/2,n]$) remain open in general.

3) ATTEMPTED SOLUTION

PHASE 1 (quick tests / sanity). For any fixed $k$, the consecutive block $1,2,\dots,k$ always lies inside $\{1,\dots,n\}$ once $n\ge k$. If additionally $n^{\varepsilon}\ge k$, then every prime factor of each $m\in\{1,\dots,k\}$ is at most $m\le k\le n^{\varepsilon}$, so the entire block is $n^{\varepsilon}$-smooth.

PHASE 2 (write a complete proof). Choose $n_0(\varepsilon,k)$ so that $n_0\ge k$ and $n_0^{\varepsilon}\ge k$; then for all $n\ge n_0$ the block $\{1,\dots,k\}$ works.

4) DETAILED WORK (with complete reasoning)

Let $\varepsilon>0$ and $k\ge 2$ be fixed.

Define \[ n_0(\varepsilon,k) \coloneqq \left\lceil \max\left\{k,\; k^{1/\varepsilon}\right\} \right\rceil. \] Then $n_0(\varepsilon,k)\ge k$ and $n_0(\varepsilon,k)^{\varepsilon}\ge k$.

Now fix any $n\ge n_0(\varepsilon,k)$. Consider the $k$ consecutive integers \[ 1,2,\dots,k. \] Because $n\ge k$, we have $\{1,\dots,k\}\subseteq\{1,\dots,n\}$.

It remains to prove that each $m\in\{1,\dots,k\}$ is $n^{\varepsilon}$-smooth. Let such an $m$ be given, and let $p$ be any prime divisor of $m$. Then $p\le m\le k$. Since $n\ge n_0(\varepsilon,k)\ge k^{1/\varepsilon}$, we have $n^{\varepsilon}\ge k$. Hence \[ p \le k \le n^{\varepsilon}. \] Because every prime divisor $p$ of $m$ satisfies $p\le n^{\varepsilon}$, the integer $m$ is $n^{\varepsilon}$-smooth.

Therefore $1,2,\dots,k$ is a block of $k$ consecutive integers in $\{1,\dots,n\}$, all $n^{\varepsilon}$-smooth. This holds for every $n\ge n_0(\varepsilon,k)$.

5) VERIFICATION

We check all requirements for the literal statement.

• For $n\ge n_0$, the chosen block $\{1,\dots,k\}$ is contained in $\{1,\dots,n\}$ because $n_0\ge k$.
• For each $m\in\{1,\dots,k\}$, every prime divisor satisfies $p\le m\le k\le n^{\varepsilon}$ because $n\ge k^{1/\varepsilon}$.
• Edge cases: if $k=2$ the same proof works; if $\varepsilon$ is small, $k^{1/\varepsilon}$ is large but finite, so $n_0$ exists.

6) FINAL

FULL SOLUTIONFULL PROOF

7) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 100%.