gpt pro 5.2 attempt on Math-5368, version 2

Round 2 $v2$: Infinite family of even integers not $3$-representable

1) ROUND-2 OBJECTIVE

Path (C): obstruction / correction. We prove a structural obstruction strictly beyond Round 1: there are infinitely many even integers that are not $3$-representable. Consequently, any universal constant $k$ (if it exists) must satisfy $k\ge 4$.

2) ROUND-1 FOUNDATION USED

We use the following Round 1 results (without re-proving them):

• Lemma 10.1 (binary-weight reformulation). An integer $n$ is $k$-representable iff there exists $m\ge 0$ with $\mathrm{wt}_2(m)\le k$ such that $n-m$ is prime.
• Lemma 10.2 (parity reduction for even targets). If $n$ is even and $n=p+S$ with $p$ an odd prime and $\mathrm{wt}_2(S)\le k$, then $S=1+m$ with $m$ even and $\mathrm{wt}_2(m)\le k-1$, so $n-1=p+m$.

3) NEW INSIGHT / TOOL (ROUND-2)

New external input: Crocker $1971$. Crocker constructs infinitely many odd integers $t$ with

1. $t\equiv -1 \pmod{16}$,
2. $t$ is not representable as $p+2^a$ with $p$ prime and $a\ge 1$,
3. $t$ is not representable as $p+2^a+2^b$ with $p$ prime and $a,b\ge 1$.

(We use this as a cited theorem; see References.)

New local lemma (proved here). From the congruence $t\equiv -1\pmod{16}$ we deduce a Hamming-weight lower bound $\mathrm{wt}_2(t-1)\ge 4$ for all $t>15$, which blocks any $k=3$ representation using the even prime $2$.

4) ATTACK PLAN (ROUND-2)

Round 1 gap. Round 1 verified one explicit even $k=3$ counterexample, but had no infinite mechanism.

Plan. Take $t$ from Crocker's infinite family and set $n=t+1$ (even). If $n$ had a $3$-representation $n=p+m$ with $\mathrm{wt}_2(m)\le 3$, then:

• if $p$ is odd, Lemma 10.2 forces $t=n-1$ to be representable as $p+m'$ with $m'$ even and $\mathrm{wt}_2(m')\le 2$, i.e. $m'\in\{0,2^a,2^a+2^b\}$ with $a,b\ge 1$; Crocker forbids this;
• if $p=2$, then $m=n-2=t-1$ and we would need $\mathrm{wt}_2(t-1)\le 3$, contradicted by the new Hamming-weight lemma coming from $t\equiv -1\pmod{16}$.

Hence each such $n$ is a $k=3$ counterexample, giving infinitely many.

5) WORK (ROUND-2)

5.1 Congruence-to-weight lemma

Lemma 5.1. Let $r\ge 1$. If $t\equiv -1\pmod{2^r}$ and $t>2^r-1$, then \[ \mathrm{wt}_2(t-1)\ \ge\ r. \] In particular, if $t\equiv -1\pmod{16}$ and $t>15$, then $\mathrm{wt}_2(t-1)\ge 4$.

Proof. Write $t=2^r q+(2^r-1)$ with an integer $q\ge 1$. Then \[ t-1=2^r q+(2^r-2). \] The number $2^r-2$ has binary form $(111\cdots 110)_2$ (with exactly $r-1$ ones), and $2^r q$ has its lowest $r$ bits equal to $0$, so there is no carry interaction when adding these two summands. Therefore \[ \mathrm{wt}_2(t-1)=\mathrm{wt}_2(2^r q)+\mathrm{wt}_2(2^r-2)=\mathrm{wt}_2(q)+(r-1)\ge 1+(r-1)=r. \] $\square$

5.2 Infinite family of even integers not $3$-representable

Theorem 5.2 (Crocker, 1971; extracted form). There exist infinitely many odd integers $t>15$ such that:

1. $t\equiv -1\pmod{16}$;
2. $t\neq p+2^a$ for any prime $p$ and any integer $a\ge 1$;
3. $t\neq p+2^a+2^b$ for any prime $p$ and integers $a,b\ge 1$.

Theorem 5.3 (Round 2 main theorem). There exist infinitely many even integers $n$ that are not $3$-representable.

Proof. Let $t$ be an odd integer satisfying Theorem 5.2, and set $n:=t+1$ (even). Assume for contradiction that $n$ is $3$-representable. By Lemma 10.1 there exists $m\ge 0$ with $\mathrm{wt}_2(m)\le 3$ such that $p:=n-m$ is prime, i.e. \[ n=p+m,\qquad \mathrm{wt}_2(m)\le 3. \] We split into cases.

Case 1: $p$ is an odd prime. Then $m=n-p$ is odd. Apply Lemma 10.2 with $k=3$ to write $m=1+m'$ where $m'$ is even and $\mathrm{wt}_2(m')\le 2$. Hence \[ t=n-1=(p+m)-1=p+m'. \] Because $m'$ is even and $\mathrm{wt}_2(m')\le 2$, we have $m'\in\{0,2^a,2^a+2^b\}$ with $a,b\ge 1$ (distinct in the last option). This contradicts Theorem 5.2.

Case 2: $p=2$. Then $m=n-2=t-1$, so $\mathrm{wt}_2(t-1)\le 3$. But $t\equiv -1\pmod{16}$ and $t>15$, so Lemma 5.1 (with $r=4$) gives $\mathrm{wt}_2(t-1)\ge 4$, contradiction.

Both cases contradict, hence $n$ is not $3$-representable. Since Theorem 5.2 provides infinitely many such $t$, we obtain infinitely many even counterexamples $n=t+1$. $\square$

5.3 Consequence for the original question

Corollary 5.4. If there exists an absolute $k$ such that every integer $n\ge 2$ is $k$-representable, then necessarily $k\ge 4$.

Proof. Theorem 5.3 gives infinitely many integers that are not $3$-representable, so no universal $k\le 3$ can work. $\square$

6) ADVERSARIAL VERIFICATION

• Repetitions of powers of $2$. Eliminated by Lemma 10.1: allowing repetition is equivalent to a Hamming-weight bound.
• The $p=2$ edge case. Handled separately via Lemma 5.1. Note that the exceptional small value $t=15$ would fail Lemma 5.1, but Crocker's family contains infinitely many $t>15$.
• $m=0$ possibility. For $n=t+1>16$ even, $m=0$ would force $p=n$ an even prime, impossible (only $2$ is even prime).
• Use of Crocker. We only use the three listed properties (including the imposed congruence $t\equiv -1\!\!\pmod{16}$) and the fact that infinitely many such $t$ exist.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED). We proved (using Round 1 Lemmas 10.1--10.2 and Crocker $1971$) that there are infinitely many even integers not $3$-representable. Hence any universal $k$ must satisfy $k\ge 4$. The existence of a universal $k$ remains open.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 45%

9) REFERENCES

1. R. Crocker, On the sum of a prime and of two powers of two, Pacific J. Math. 36 $1971$, 103--107.