gpt pro 5.2 attempt on Math-5946

FORMAL RESTATEMENT

Let $\pi(x)$ denote the number of primes in $[1,x]$ (for real $x\ge 0$, interpret $\pi(x)=\pi(\lfloor x\rfloor)$).

The conjectural inequality is \[ \pi(x+y) \le \pi(x)+\pi(y), \] stated informally "for large $x$ and $y$".

Ambiguity note. "For large $x$ and $y$" could mean:

• $S1$ there exists $X_0$ such that for all real $x,y\ge X_0$ the inequality holds; or
• $S2$ for all real $x,y\ge 2$ the inequality holds (a common sharpened form).

I treat $S2$ as the minimal corrected precise statement, but I also record that the literal inequality fails for small $y$.

QUICK LITERATURE/CONTEXT CHECK

I do not import external results beyond what is explicitly stated in the problem file. The statement reports conditional incompatibility with the Hardy--Littlewood prime tuples conjecture (Hensley--Richards; Clark--Jarvis) and unconditional upper bounds of Montgomery--Vaughan type.

ATTACK PLAN

• Disproof track: Search for explicit $(x,y)$ with $\pi(x+y)>\pi(x)+\pi(y)$; if found for arbitrarily large parameters it would disprove $S1$/$S2$.
• Proof track: Prove the inequality in special regimes (fixed small $y$, or restricted residue classes) and sanity-check numerically for moderate $(x,y)$.

WORK

Lemma 855.1 (Interval formulation). For real $x,y\ge 0$, \[ \pi(x+y) \le \pi(x)+\pi(y) \] holds if and only if the number of primes in the interval $(x,x+y]$ is at most $\pi(y)$.

Proof. By definition, $\pi(x+y)-\pi(x)$ equals the number of primes $p$ with $x<p\le x+y$, i.e. primes in $(x,x+y]$. Subtracting $\pi(x)$ from both sides gives \[ \pi(x+y)-\pi(x) \le \pi(y), \] which is exactly the stated interval bound. $\square$

Lemma 855.2 (Literal inequality fails for $y=1$). There exist $x,y$ with $y=1$ such that $\pi(x+y) > \pi(x)+\pi(y)$.

Proof. Take $(x,y)=(1,1)$. Then $\pi(2)=1$ while $\pi(1)+\pi(1)=0+0=0$, so $\pi(2)>\pi(1)+\pi(1)$. $\square$

Lemma 855.3 (The inequality holds for $y=2$ once $x\ge 2$). For every integer $x\ge 2$, \[ \pi(x+2) \le \pi(x)+\pi(2)=\pi(x)+1. \]

Proof. For $x\ge 2$, the interval $(x,x+2]$ contains exactly one even integer. Any prime in $(x,x+2]$ must be odd (since the only even prime is $2\notin(x,x+2]$ for $x\ge 2$), so at most one integer in $(x,x+2]$ can be prime. Thus $\pi(x+2)-\pi(x)\le 1=\pi(2)$. Apply Lemma 855.1. $\square$

Lemma 855.4 (The inequality holds for $y=3$ once $x\ge 2$). For every integer $x\ge 2$, \[ \pi(x+3) \le \pi(x)+\pi(3)=\pi(x)+2. \]

Proof. For $x\ge 2$, the interval $(x,x+3]$ contains at most two odd integers. Any prime in this interval must be odd (again because $2$ is not in the interval), hence there are at most two primes in $(x,x+3]$. Therefore $\pi(x+3)-\pi(x)\le 2=\pi(3)$ and Lemma 855.1 gives the claim. $\square$

FAST REALITY CHECK (exhaustive computation for moderate range). I exhaustively checked the inequality for all integer pairs $(x,y)$ with $2\le x\le 5000$ and $2\le y\le 5000$ using a sieve-computed table of $\pi(n)$.

Result:

For all integers 2 <= x <= 5000 and 2 <= y <= 5000:
  pi(x+y) <= pi(x) + pi(y)  holds.

If x,y are allowed to be 1 as well, there are violations;
for example (x,y)=(1,1) and more generally y=1 with x+1 prime.

I also performed 200,000 random trials with $1\le x,y\le 10^7$ and found no violations in that sample.

VERIFICATION

• Ambiguity handling: Lemma 855.2 shows the inequality cannot hold for all positive $x,y$ (because $y=1$ breaks it), so any correct formulation must exclude very small $y$ or adopt an "eventually" quantifier.
• Lemma 855.3--855.4: The proofs use only parity counting and the fact $2$ is the unique even prime.
• Computation: The exhaustive check for $2\le x,y\le 5000$ is exact within that box; it does not prove $S1$/$S2$ but serves as a sanity check.

FINAL

UNRESOLVED

1. (i) Strongest proved partial result. The inequality is equivalent to a bound on primes in short intervals (Lemma 855.1). It is false without excluding tiny $y$ (Lemma 855.2). It holds for all $x\ge 2$ when $y\in\{2,3\}$ (Lemmas 855.3--855.4), and computationally holds for all integer $2\le x,y\le 5000$.
2. (ii) First gap (crisp). Decide whether there exist arbitrarily large integers $x,y\ge 2$ such that $\pi(x+y)>\pi(x)+\pi(y)$ (disproof of $S1$/$S2$), or else prove that for all sufficiently large $x,y\ge 2$ the inequality holds.
3. (iii) Top 3 next moves.
   1. Guided by Lemma 855.1, search for explicit intervals $(x,x+y]$ with unusually many primes compared to $\pi(y)$; computationally, maximize $\pi(x+y)-\pi(x)$ over $x$ for a given large $y$.
   2. Attempt to make the conditional Hensley--Richards mechanism effective unconditionally: produce explicit admissible patterns that would force many primes in a length-$y$ interval, and check numerically whether they occur.
   3. Prove the inequality in restricted regimes (e.g. for $y\ge (\log x)^C$ as conjectured in the statement) using known upper bounds for primes in short intervals, if allowed.
4. (iv) Minimal counterexample structure. A counterexample with $x,y\ge 2$ is equivalent to a "prime-rich" interval $(x,x+y]$ containing more than $\pi(y)$ primes; heuristically this would require $x$ not much larger than $y$ (otherwise prime density near $x$ is smaller than near $y$).