gpt pro 5.4 attempt on Math-794

Problem 389. A divisibility problem for two consecutive blocks

1) ROUND-2 OBJECTIVE

Path (A) was not realistic after Round 1, because no general existence theorem for all $n$ had been found. The most promising Round-2 goal is therefore the fail-safe route: prove theorems that strictly advance beyond Round 1.

I will do this in three ways.

• I derive an explicit digit-sum criterion from the Round-1 valuation reformulation.
• I give infinite families of solutions for $n=2$ and $n=3$ (and the trivial complete solution for $n=1$).
• I prove that for every fixed $n$, infinitely many $k$ fail.

So the global problem remains open, but the picture is now much sharper.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts.

• The problem asks whether for every $n\ge 1$ there exists $k\ge 1$ such that \[ \prod_{i=0}^{k-1}(n+i)\mid \prod_{i=0}^{k-1}(n+k+i). \]

• Round-1 binomial reformulation: \[ \prod_{i=0}^{k-1}(n+i)\mid \prod_{i=0}^{k-1}(n+k+i) \quad\Leftrightarrow\quad {n+k-1\choose k}\mid {n+2k-1\choose k}. \]

• Round-1 valuation reformulation: the divisibility is equivalent to \[ v_p\!\left((n+2k-1)!\right)+v_p\!\left((n-1)!\right)\ge 2v_p\!\left((n+k-1)!\right) \] for every prime $p$.

3) NEW INSIGHT / TOOL (ROUND-2)

The key new tool is a digit-sum criterion obtained from the Round-1 valuation identity via Legendre's formula.

If $s_p(m)$ denotes the sum of the base-$p$ digits of $m$, then \[ v_p(t!)=\frac{t-s_p(t)}{p-1}. \] Substituting this into the Round-1 valuation condition yields an exact criterion in terms of \[ 2s_p(n+k-1)-s_p(n+2k-1)-s_p(n-1). \]

This criterion is strong enough to prove:

• every $k$ works for $n=1$;
• infinitely many $k$ work for $n=2$;
• infinitely many $k$ work for $n=3$;
• for each fixed $n$, infinitely many $k$ do not work.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was the absence of any constructive mechanism producing or ruling out admissible $k$.

To overcome that, I first derive the digit-sum test. Then I use it in two opposite directions:

1. choose $k$ so that the relevant base-$p$ digit sums force divisibility (positive families);
2. choose $k$ so that one particular prime gives a negative digit-sum expression (negative families).

This supplies strict, rigorous progress on both the existence and nonexistence sides.

5) WORK (ROUND-2)

Define \[ T_n:=\left\{k\ge 1:\ \prod_{i=0}^{k-1}(n+i)\mid \prod_{i=0}^{k-1}(n+k+i)\right\}. \]

Claim 1 (digit-sum criterion). A positive integer $k$ belongs to $T_n$ if and only if for every prime $p$ dividing $\prod_{i=0}^{k-1}(n+i)$ one has \[ 2s_p(n+k-1)-s_p(n+2k-1)-s_p(n-1)\ge 0. \]

Proof. By Round-1, $k\in T_n$ if and only if \[ v_p\!\left((n+2k-1)!\right)+v_p\!\left((n-1)!\right)-2v_p\!\left((n+k-1)!\right)\ge 0 \] for every prime $p$.

Now apply Legendre's formula \[ v_p(t!)=\frac{t-s_p(t)}{p-1}. \] The left-hand side becomes \[ \frac{(n+2k-1)-s_p(n+2k-1)+(n-1)-s_p(n-1)-2(n+k-1)+2s_p(n+k-1)}{p-1}, \] which simplifies to \[ \frac{2s_p(n+k-1)-s_p(n+2k-1)-s_p(n-1)}{p-1}. \] Therefore the valuation inequality is equivalent to \[ 2s_p(n+k-1)-s_p(n+2k-1)-s_p(n-1)\ge 0. \]

It is enough to check primes dividing the left block $\prod_{i=0}^{k-1}(n+i)$, because if a prime $p$ does not divide that block, then its left-side valuation is $0$ and the divisibility condition at $p$ is automatic. QED.

Claim 2. $T_1=\{1,2,3,\dots\}$. In other words, every $k\ge 1$ works for $n=1$.

Proof. For $n=1$ the divisibility becomes \[ 1\cdot 2\cdots k \mid (k+1)(k+2)\cdots(2k). \] The left-hand side is $k!$, and the right-hand side is \[ \frac{(2k)!}{k!}={2k\choose k}\,k!. \] Hence the right-hand side is an integer multiple of $k!$, so every $k$ works. QED.

Claim 3. $T_2$ is infinite. More precisely, if $p>5$ and $q$ are primes with \[ \frac{3p}{2}<q<2p, \] then \[ k=pq-1 \] belongs to $T_2$.

Proof. For $n=2$, Round-1 gives \[ k\in T_2 \quad\Leftrightarrow\quad {k+1\choose k}=k+1\mid {2k+1\choose k}. \] So it suffices to show that $k+1=pq$ divides ${2k+1\choose k}$.

Because $q<2p$, the prime factorization of $k+1$ is exactly $pq$ with two distinct primes $p$ and $q$. Hence it is enough to show that both $p$ and $q$ divide ${2k+1\choose k}$.

By Claim 1 with $n=2$, it is enough to verify \[ 2s_r(k+1)-s_r(2k+1)-1\ge 0 \] for $r=p$ and $r=q$.

Base $p$. Since $k+1=pq$ and $q<2p$, we can write \[ k+1=pq=p^2+(q-p)p. \] Therefore \[ s_p(k+1)=1+(q-p). \] Also \[ 2k+1=2pq-1=3p^2+(2q-3p-1)p+(p-1). \] The coefficient $2q-3p-1$ is a valid base-$p$ digit because $q>3p/2$ implies $2q-3p\ge 1$, and $q<2p$ implies $2q-3p-1\le p-2$. Hence \[ s_p(2k+1)=3+(2q-3p-1)+(p-1)=2q-2p+1. \] Thus \[ 2s_p(k+1)-s_p(2k+1)-1 =2(q-p+1)-(2q-2p+1)-1=0. \]

Base $q$. Now \[ k+1=pq=p\cdot q+0, \] so \[ s_q(k+1)=p. \] Also \[ 2k+1=2pq-1=q^2+(2p-q-1)q+(q-1). \] Again the coefficient $2p-q-1$ is a valid base-$q$ digit because $q<2p$ implies $2p-q-1\ge 0$, and $q>p$ implies $2p-q-1<q$. Therefore \[ s_q(2k+1)=1+(2p-q-1)+(q-1)=2p-1, \] and hence \[ 2s_q(k+1)-s_q(2k+1)-1=2p-(2p-1)-1=0. \]

So both primes $p$ and $q$ divide ${2k+1\choose k}$, hence $pq=k+1$ divides ${2k+1\choose k}$, and therefore $k\in T_2$.

Finally, there are infinitely many such prime pairs $(p,q)$. Indeed, Nagura's theorem gives a prime in $(x,6x/5)$ for every sufficiently large $x$; taking $x=3p/2$ yields a prime \[ q\in\left(\frac{3p}{2},\frac{9p}{5}\right)\subset\left(\frac{3p}{2},2p\right) \] for every sufficiently large prime $p$. So $T_2$ is infinite. QED.

Claim 4. $T_3$ is infinite. More precisely, if $k\in T_2$, then $k-1\in T_3$.

Proof. Assume $k\in T_2$. Then by the $n=2$ binomial reformulation, \[ k+1\mid {2k+1\choose k}. \] Use the standard identity \[ {2k+1\choose k}=\frac{2k+1}{k+1}{2k\choose k}. \] Multiplying the divisibility $k+1\mid {2k+1\choose k}$ by $k$ gives \[ k(k+1)\mid k{2k+1\choose k}=(2k+1){2k\choose k-1}. \] Now \[ \gcd(2k+1,k)=1,\qquad \gcd(2k+1,k+1)=1, \] so \[ \gcd(2k+1,k(k+1))=1. \] Therefore \[ k(k+1)\mid {2k\choose k-1}. \] Since $k(k+1)$ is always even, this implies \[ {k+1\choose 2}=\frac{k(k+1)}{2}\mid {2k\choose k-1}. \] Now apply the Round-1 binomial reformulation with $n=3$ and $K:=k-1$: \[ K\in T_3 \quad\Leftrightarrow\quad {K+2\choose K}={K+2\choose 2}\mid {2K+2\choose K}. \] Substituting $K=k-1$ gives exactly \[ {k+1\choose 2}\mid {2k\choose k-1}, \] which we have proved. Hence $k-1\in T_3$. QED.

Combining Claims 3 and 4, the explicit family \[ k=pq-2 \] belongs to $T_3$ whenever $p>5$ and $q$ are primes with $3p/2<q<2p$.

Claim 5. For every fixed $n\ge 2$, the complement of $T_n$ is infinite. In fact, if $p>n-1$ is prime and $m\ge 2$, then \[ k=2p^m-n+1 \] does not belong to $T_n$.

Proof. Fix $n\ge 2$, choose a prime $p>n-1$, choose $m\ge 2$, and define \[ k=2p^m-n+1. \] Then \[ n+k-1=2p^m, \] so $p$ divides the left block $\prod_{i=0}^{k-1}(n+i)$, and Claim 1 applies to this prime $p$.

Because $p>n-1$, the base-$p$ expansion of $n-1$ is a single digit, so \[ s_p(n-1)=n-1. \] Also \[ n+k-1=2p^m, \] hence \[ s_p(n+k-1)=2. \]

Now \[ n+2k-1=4p^m-n+1. \] Write this in base $p$ as \[ 4p^m-n+1 =3p^m+(p-1)(p+p^2+\cdots+p^{m-1})+(p-n+1). \] This is valid because $p>n-1$, so $0\le p-n+1\le p-1$. Therefore \[ s_p(n+2k-1)=3+(m-1)(p-1)+(p-n+1). \]

Insert these values into Claim 1: \[ 2s_p(n+k-1)-s_p(n+2k-1)-s_p(n-1) \] \[ =4-\left(3+(m-1)(p-1)+(p-n+1)\right)-(n-1) =1-p-(m-1)(p-1)<0. \] So the digit-sum criterion fails for the prime $p$, and therefore $k\notin T_n$.

Since there are infinitely many choices of $p$ and $m$, the complement of $T_n$ is infinite. QED.

6) ADVERSARIAL VERIFICATION

I check the new arguments against the main possible flaws.

• Index shift. The literature on this problem often uses the parameter $n-1$ instead of $n$. To avoid any hidden mismatch, every statement above was re-derived in the present notation.

• Claim 1 uses only primes dividing the left block. That is correct: divisibility needs to be checked only at primes that occur on the left. For primes not dividing the left block, the left valuation is $0$ and the divisibility condition is automatic.

• Base-$p$ digit expansions in Claim 3. The inequalities \[ \frac{3p}{2}<q<2p \] were used exactly to ensure that the coefficients in the base-$p$ and base-$q$ expansions are valid digits. I checked both bounds explicitly in the proof.

• Infinitely many prime pairs $(p,q)$. This does not follow from Bertrand alone, but it does follow from Nagura's theorem on primes in short intervals, which is why I used Nagura explicitly.

• The reduction from $T_2$ to $T_3$. The step \[ k(k+1)\mid (2k+1){2k\choose k-1} \quad\Rightarrow\quad k(k+1)\mid {2k\choose k-1} \] uses $\gcd(2k+1,k(k+1))=1$, and this was checked directly.

• Claim 5. The chosen prime $p$ really does divide the left block because $n+k-1=2p^m$ is the last term of that block. The base-$p$ expansion of $4p^m-n+1$ was checked by direct summation.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advances beyond Round 1 are the following.

1. There is an exact digit-sum criterion for membership in $T_n$.

2. The problem is completely solved for $n=1$: every $k$ works, so in this case the complement of $T_1$ is empty.

3. There are infinitely many solutions for $n=2$, for example \[ k=pq-1 \] with primes $p>5$ and $3p/2<q<2p$.

4. There are infinitely many solutions for $n=3$, for example \[ k=pq-2 \] with the same prime-pair condition.

5. For every fixed $n\ge 2$, there are infinitely many failures; indeed \[ k=2p^m-n+1 \] never works whenever $p>n-1$ is prime and $m\ge 2$.

What remains open is the original global existence question: for each $n\ge 1$, must the set $T_n$ be nonempty?

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 45%

9) REFERENCES

1. M. Ulas, On the Erdős--Straus and Erdős--Graham problems, especially the digit-sum criterion and the explicit families for the shifted indexing version of this problem.

2. J. Nagura, On the interval containing at least one prime number, for the prime-in-short-interval statement used to make the family in Claim 3 infinite.

3. The Erdős Problems site, Problem 389, and OEIS A375071 for computational background only.