Erdős Problem #383

Problem #383 (Round 2)

1) ROUND-2 OBJECTIVE

Path A (proof-oriented constructive progress). Round 1 had examples through $k=6$ and no example for $k=7$ in the search range used there. The new aim is to push the exact search further and replace "none found" by explicit rigorous constructions for larger $k$.

2) Round-1 FOUNDATION USED

I use the Round-1 reformulation:

for a prime $p>k$, \[ P\!\left(\prod_{i=0}^{k}(p^2+i)\right)=p \] is equivalent to \[ P(p^2+i)\le p\qquad (1\le i\le k). \]

So it is enough to exhibit primes $p$ such that each of the integers \[ p^2+1,\dots,p^2+k \] is $p$-smooth.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is not a new theorem but a new exact computational certificate: I found and verified explicit primes for \[ k=7,\ 8,\ 9. \] This strictly advances Round 1, which only had examples through $k=6$.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was the absence of any example for $k=7$ in the earlier search range. The new plan is:

1. search much farther than Round 1;
2. for each successful prime $p$, give exact factorisations of $p^2+i$ for all $1\le i\le k$;
3. conclude rigorously that the largest prime factor of the whole product is exactly $p$.

5) WORK (ROUND-2)

Example for $k=7$.

Take \[ p=510569. \] Then the exact factorisations are:

1. $p^2+1=2\cdot 37\cdot 31721\cdot 111053$;
2. $p^2+2=3\cdot 2731\cdot 4651\cdot 6841$;
3. $p^2+3=2^2\cdot 13\cdot 43\cdot 523\cdot 222913$;
4. $p^2+4=5\cdot 17\cdot 29^2\cdot 641\cdot 5689$;
5. $p^2+5=2\cdot 3\cdot 7\cdot 23\cdot 167\cdot 1061\cdot 1523$;
6. $p^2+6=11\cdot 59\cdot 101\cdot 991\cdot 4013$;
7. $p^2+7=2^3\cdot 71\cdot 401\cdot 449\cdot 2549$.

Every prime shown is $<p$. Since $p>7$, none of $p^2+1,\dots,p^2+7$ is divisible by $p$. Therefore the factor $p^2$ contributes the only copies of $p$ to \[ \prod_{i=0}^{7}(p^2+i), \] and all other prime divisors are $<p$. Hence \[ P\!\left(\prod_{i=0}^{7}(p^2+i)\right)=p. \]

Example for $k=8$.

Take \[ p=2869123. \] Then:

1. $p^2+1=2\cdot 5\cdot 489337\cdot 1682249$;
2. $p^2+2=3^6\cdot 82129\cdot 137491$;
3. $p^2+3=2^2\cdot 7\cdot 31\cdot 5449\cdot 1740451$;
4. $p^2+4=13\cdot 251969\cdot 2513089$;
5. $p^2+5=2\cdot 3\cdot 43\cdot 563\cdot 1321\cdot 42901$;
6. $p^2+6=5\cdot 11^2\cdot 149\cdot 313\cdot 291751$;
7. $p^2+7=2^4\cdot 701\cdot 9473\cdot 77477$;
8. $p^2+8=3\cdot 953\cdot 4003\cdot 719281$.

Again every displayed prime is $<p$, and $p>8$ implies $p$ does not divide $p^2+i$ for $1\le i\le 8$. Therefore \[ P\!\left(\prod_{i=0}^{8}(p^2+i)\right)=p. \]

Example for $k=9$.

Take \[ p=3092701. \] Then:

1. $p^2+1=2\cdot 1373\cdot 7237\cdot 481301$;
2. $p^2+2=3^2\cdot 17\cdot 41\cdot 89\cdot 1289\cdot 13291$;
3. $p^2+3=2^2\cdot 73\cdot 104947\cdot 312121$;
4. $p^2+4=5\cdot 101\cdot 137\cdot 1553\cdot 89021$;
5. $p^2+5=2\cdot 3\cdot 7^2\cdot 30781\cdot 1056929$;
6. $p^2+6=11\cdot 1709\cdot 5021\cdot 101333$;
7. $p^2+7=2^4\cdot 919\cdot 2417\cdot 269131$;
8. $p^2+8=3\cdot 641\cdot 5939\cdot 837497$;
9. $p^2+9=2\cdot 5\cdot 37\cdot 15313\cdot 1688161$.

All displayed prime divisors are $<p$, and $p>9$ implies $p$ does not divide $p^2+i$ for $1\le i\le 9$. Hence \[ P\!\left(\prod_{i=0}^{9}(p^2+i)\right)=p. \]

Consequences for Problem #382.

Each of the three examples above also gives a bad interval: \[ [p^2,p^2+7],\qquad [p^2,p^2+8],\qquad [p^2,p^2+9]. \] So the Problem #383 search now yields rigorous constructions for bad intervals of lengths $7,8,9$.

6) ADVERSARIAL VERIFICATION

I checked the key logical points.

1. Could one of the numbers $p^2+i$ be divisible by $p$? No, because $1\le i\le k<p$ in all three examples, so $p$ does not divide $p^2+i$.
2. Is it enough to factor only $p^2+1,\dots,p^2+k$? Yes. The factor $p^2$ already contributes the prime $p$, and once all other factors are shown to have prime divisors $<p$, the largest prime divisor of the full product is exactly $p$.
3. Could there be a larger hidden prime factor in one of the listed numbers? No. The displayed decompositions are exact factorisations.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 replaces the Round-1 frontier \[ k\le 6 \] by the rigorously verified frontier \[ k\le 9, \] through explicit primes for \[ k=7,\ 8,\ 9. \]

This is substantial constructive progress toward the conjecture, but it does not prove that for every fixed $k$ there are infinitely many such primes.

8) COMPLETION ESTIMATE

COMPLETION: 35%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #383, accessed 2026-03-08, https://www.erdosproblems.com/383.

Problem 383: auditing smooth intervals and their immediate endpoints

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. The target described in 383.tex, read completely, asks for arbitrarily long, and ultimately infinitely many for each fixed length, intervals beginning at a prime square with largest prime factor equal to that prime. The source supplies certificates through length nine. I recheck all of them and try the immediate extension of each certificate. This produces verified boundaries, not a new length-ten example.

2) WORK

For each source pair $(p,k)=(510569,7),(2869123,8),(3092701,9)$, the verification program multiplies every displayed factor list, tests $p$ and every listed factor for primality by exhaustive trial division up to its integer square root, and checks that all factors of $p^2+i$, $1\le i\le k$, are less than $p$. All 24 displayed factorizations pass. Thus all three source certificates are valid.

Factoring the next integer shows that none of these particular primes gives a longer interval: \[ \begin{split} 510569^2+8&=3^2\cdot419\cdot69127739,\\ 2869123^2+9&=2\cdot4115933394569,\\ 3092701^2+10&=23\cdot211\cdot1970904487. \end{split} \] All displayed factors in these three new factorizations were tested prime by the same deterministic method, and in each row the largest factor exceeds $p$.

A simple necessary filter for future searches is worth making explicit. If $p^2+i$ is $p$-smooth for any $i>0$, then its number $\Omega(p^2+i)$ of prime factors counted with multiplicity is at least three. Otherwise it is the product of at most two primes at most $p$ and hence is at most $p^2$, a contradiction. Moreover, for odd $p$, every odd prime divisor $q$ of $p^2+1$ satisfies $q\equiv1\pmod4$: the nonzero residue $p\bmod q$ squares to $-1$, so its multiplicative order is four and divides $q-1$. These are necessary filters, not sufficient smoothness criteria.

3) ADVERSARIAL VERIFICATION

Checking factor products alone would not certify the claimed prime factors; explicit primality checks were also executed. Conversely, the primality of the large next-step factors is needed to exclude smoothness, and was not assumed from their size. The interval from $p^2$ through $p^2+k$ contains $k+1$ integers, so the convention here counts the $k$ increments after the square. A single example for a fixed $k$ does not establish infinitude for that $k$.

4) FINAL

UNRESOLVED. The supplied frontier $k=9$ is independently certified, and each supplied example's first failure is explicit. No verified improvement of that frontier or infinitude theorem was obtained.

5) SOURCES CHECKED

The complete supplied 383.tex; all factor lists and tests are in verification/batch_1_checks.py and its JSON output, executed 2026-09-23. The two search filters are proved directly above.

6) COMPLETION ESTIMATE

Subjective progress toward the full arbitrary-length and infinitude target. COMPLETION: 10%

Problem #383

1) FORMAL RESTATEMENT

Fix an integer $k\ge 0$. For a prime $p$, define \[ F_k(p)\ :=\ \prod_{i=0}^{k}(p^2+i). \] Let $P(n)$ denote the largest prime factor of $n$ (with $P(1):=1$).

Question. Is it true that for every fixed $k$ there exist infinitely many primes $p$ such that \[ P\left(F_k(p)\right) = p\ ? \]

Equivalently (for $p>k$ so that $p\nmid p^2+i$ for $1\le i\le k$), the condition is: \[ \max_{1\le i\le k} P(p^2+i)\ \le\ p. \] Indeed, the factor $p^2$ ensures that $p$ divides $F_k(p)$, and the requirement that no prime $>p$ divides any of the other factors forces $p$ to be the largest prime divisor.

2) QUICK LITERATURE/CONTEXT CHECK

The Erdős Problems forum thread for #383 lists the problem as open and notes that a positive answer would imply the second part of Problem #382.(https://www.erdosproblems.com/forum/thread/383 (accessed 2026-01-17).) It also gives a heuristic: the "probability" that an integer $n$ has no prime divisor $\ge n^{1/2}$ is $1-\log 2>0$, suggesting the answer should be yes.(Same source as above.) A comment in the thread mentions a conditional approach for the first nontrivial case using conjectures about Newman--Shanks--Williams primes.(Same source as above; comment by user "Dogmachine".)

3) ATTACK PLAN

1. Re-express the condition as a simultaneous smoothness requirement: $p^2+i$ must be $p$-smooth for $1\le i\le k$.
2. Produce computational evidence for small $k$ by searching primes $p$ up to various cutoffs.
3. Heuristically, model the events "$p^2+i$ is $p$-smooth" as roughly independent with probability $\approx 1-\log 2$; then the expected count up to $X$ is $\asymp \frac{X}{\log X}(1-\log 2)^k$, which diverges for fixed $k$.
4. The hard step is making any of this unconditional: it is a problem about smooth values of quadratic polynomials at prime arguments, well beyond current sieve technology in general.

4) WORK

Computational evidence.

For each fixed $k$, we searched primes $p$ up to a cutoff $P_{\max}$ and checked whether $P(p^2+i)\le p$ holds for all $1\le i\le k$. The table below records counts and the first few solutions found.

These computations support (but do not prove) the conjecture that for each fixed $k$ there may be infinitely many such primes $p$.

Heuristic.

Write $n=p^2+i\approx p^2$. The condition $P(n)\le p$ is essentially $P(n)\le n^{1/2}$. The Dickman--de Bruijn heuristic for smooth numbers suggests \[ \mathbb{P}(P(n)\le n^{1/2}) \approx \rho(2)=1-\log 2. \] Assuming (very optimistically) that the $k$ conditions for $i=1,\dots,k$ behave like independent random events of probability $1-\log 2$, one predicts that among primes $p\le X$ the expected number of solutions is \[ \sim \frac{X}{\log X}\,(1-\log 2)^k, \] which tends to infinity with $X$ for each fixed $k$.

5) VERIFICATION

• The equivalence "$P(F_k(p))=p$ iff $P(p^2+i)\le p$ for all $1\le i\le k$" holds for all primes $p>k$; for $p\le k$ one must treat separately because $p$ can divide some $p^2+i$. The computational search handled the definition directly by checking prime factors of each $p^2+i$.
• The computations above are finite and cannot distinguish "infinitely many" from "finitely many but many"; they serve only as evidence.

6) FINAL

UNRESOLVED.

1. (i) Strongest partial result obtained. Reformulated the problem as requiring simultaneous $p$-smoothness of the $k$ consecutive integers $p^2+1,\dots,p^2+k$. Provided explicit computational evidence: solutions exist for $k=1,2,3,4,5$ with many primes below $50000$, and for $k=6$ at least two primes below $200000$.
2. (ii) First hard gap. No known unconditional method currently proves that infinitely many primes $p$ make even a single quadratic value $p^2+1$ have all prime factors $\le p$ (i.e. be $p$-smooth), let alone $k$ simultaneous values.
3. (iii) Top 3 next moves.
   1. Attempt a lower-bound sieve for primes $p$ for which $p^2+i$ has no prime factor in $(p, p^{1+\varepsilon}]$ and then control larger prime factors via additional arguments (a difficult "parity problem" barrier is expected).
   2. Seek conditional proofs under strong hypotheses on prime values/smooth values of polynomials (e.g. variants of Dickson/Schinzel, or very strong distributional conjectures).
   3. Explore connections to known families where $p^2+i$ factors in a controlled way (e.g. algebraic identities or Pell-type parametrisations) that might force $p^2+i$ to be smooth for infinitely many $p$.
4. (iv) Why it might be true. Heuristics based on smooth numbers (and the constant $1-\log 2$) suggest a positive density of $p^2+i$ should have all prime factors $\le p$, and for fixed $k$ the expected number of primes $p\le X$ satisfying all $k$ constraints should still grow like $\frac{X}{\log X}(1-\log 2)^k$.

7) COMPLETION ESTIMATE

Estimated probability of a full resolution from these notes alone: 15%.