gpt pro 5.4 attempt on Math-5511

Problem 222 --- Round 2

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. The original prompt "explore the behaviour" is too broad to finish in one round. The sharpest Round 1 gap was the very weak pointwise upper bound $\Delta_k\le 2\sqrt{n_k}+1$. In this round I correct that to the classical Bambah--Chowla scale and give a complete proof: \[ \Delta_k \le 2\sqrt{2}\,(n_k+1)^{1/4}+2. \] This is a major quantitative improvement from exponent $1/2$ down to exponent $1/4$.

2) Round-1 FOUNDATION USED

I use the Round 1 notation \[ \mathcal S=\{n\ge 0:n=a^2+b^2\mbox{ for some integers }a,b\}, \qquad 0\le n_1<n_2<\cdots, \qquad \Delta_k=n_{k+1}-n_k. \] I also retain the Round 1 theorem that the gaps are unbounded (via the CRT construction), though that theorem is not reproved here.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is the arithmetic form of the Bambah--Chowla argument: for a given real $x\ge 1$, choose \[ u:=\lfloor \sqrt{x}\rfloor, \qquad v:=\left\lceil \sqrt{x-u^2}\right\rceil. \] Then $u^2+v^2$ is a sum of two squares lying very close above $x$, and a direct estimate shows the error is $O(x^{1/4})$.

4) ATTACK PLAN (ROUND-2)

The missing claim after Round 1 was a strong universal upper bound for the next representable number. So the precise target is:

Claim. For every real $x\ge 1$ there exists $m\in\mathcal S$ with \[ x\le m < x+2\sqrt{2}\,x^{1/4}+1. \]

Once this is proved, applying it with $x=n_k+1$ gives a pointwise bound for the consecutive gap $\Delta_k$.

5) WORK (ROUND-2)

Let $x\ge 1$ be real, and define \[ u:=\lfloor\sqrt{x}\rfloor, \qquad v:=\left\lceil\sqrt{x-u^2}\right\rceil. \] Then $u,v$ are nonnegative integers, $u^2\le x$, and therefore $x-u^2\ge 0$. By the defining property of the ceiling function, \[ \sqrt{x-u^2}\le v < \sqrt{x-u^2}+1. \] Squaring gives \[ x-u^2\le v^2 < x-u^2 + 2\sqrt{x-u^2}+1. \] After adding $u^2$ to all three terms we obtain \[ x\le u^2+v^2 < x + 2\sqrt{x-u^2}+1. \] So it remains to estimate $x-u^2$. Since $u=\lfloor\sqrt{x}\rfloor$, we have $u\ge \sqrt{x}-1$, hence \[ u^2 \ge (\sqrt{x}-1)^2 = x-2\sqrt{x}+1. \] Therefore \[ x-u^2 \le 2\sqrt{x}-1. \] Substituting this into the previous inequality yields \[ u^2+v^2 < x+2\sqrt{2\sqrt{x}-1}+1 < x+2\sqrt{2}\,x^{1/4}+1. \] Because $u^2+v^2\in\mathcal S$, this proves the claim.

Now take $x=n_k+1$. Then there exists $m\in\mathcal S$ with \[ n_k+1\le m < n_k+1+2\sqrt{2}\,(n_k+1)^{1/4}+1. \] Since $n_{k+1}$ is the least element of $\mathcal S$ strictly larger than $n_k$, we have $n_{k+1}\le m$. Hence \[ \Delta_k=n_{k+1}-n_k\le 2\sqrt{2}\,(n_k+1)^{1/4}+2. \] Thus Round 2 proves the following corrected theorem.

Corrected theorem proved in Round 2. For every $k\ge 1$, \[ \Delta_k \le 2\sqrt{2}\,(n_k+1)^{1/4}+2. \] Equivalently, the maximal gap function among sums of two squares satisfies \[ g(X):=\max\{n_{j+1}-n_j:n_j\le X\}\ll X^{1/4}. \] Together with the Round 1 CRT construction, this gives the rigorous window \[ g(X)\to\infty \qquad\mbox{and}\qquad g(X)\ll X^{1/4}. \]

6) ADVERSARIAL VERIFICATION

• Is $x-u^2$ always nonnegative? Yes, because $u=\lfloor\sqrt{x}\rfloor$ implies $u^2\le x$.
• Could the constructed sum of two squares equal $n_k$ when deriving the gap bound? That is why the theorem is applied to $x=n_k+1$, not to $x=n_k$.
• Any issue at small $x$? No. For example $x=1$ gives $u=1$, $v=0$, and the inequality is immediate.
• Did the exponent really drop from $1/2$ to $1/4$? Yes: the only nontrivial estimate is $x-u^2\le 2\sqrt{x}-1$, and taking one more square root produces the exponent $1/4$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED). The overall behaviour of gaps between sums of two squares is still not fully understood, but Round 2 proves a sharp classical pointwise upper bound absent from Round 1: \[ \Delta_k \le 2\sqrt{2}\,(n_k+1)^{1/4}+2. \] This closes the main quantitative gap in the Round 1 work.

8) COMPLETION ESTIMATE

COMPLETION: 65%

9) REFERENCES

• R. P. Bambah and S. Chowla, On numbers which can be expressed as a sum of two squares, Proc. Nat. Inst. Sci. India 13 $1947$, 101--103.
• P. Shiu, On a result of Bambah and Chowla, Integers 19 $2019$, A48.