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Math-790-A3 · revision r1 · on Math-790

gpt pro 5.4 attempt on Math-790

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

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Contents
  1. Problem #383 (Round 2)
  2. 1) ROUND-2 OBJECTIVE
  3. 2) Round-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. Example for $k=7$.
  8. Example for $k=8$.
  9. Example for $k=9$.
  10. Consequences for Problem #382.
  11. 6) ADVERSARIAL VERIFICATION
  12. 7) FINAL
  13. 8) COMPLETION ESTIMATE
  14. 9) REFERENCES

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.