gpt pro 5.2 attempt on Math-6115

FORMAL RESTATEMENT

Let $\sigma(n)=\sum_{d\mid n} d$. For $x\ge 1$, let $N(x)$ be the number of ordered pairs $(a,b)\in\mathbb Z_{\ge 1}^2$ such that \[ \sigma(a)+\sigma(b)=\sigma(a+b)\quad\text{and}\quad a+b\le x. \] (If one instead counts unordered pairs, the count differs by at most a factor $2$.) The question asks for the growth of $N(x)$ and in particular whether $N(x)\sim c x$ for some constant $c>0$.

QUICK LITERATURE/CONTEXT CHECK

The problem is posed as an open question (Guy B15). I do not use or claim any external results.

ATTACK PLAN

Proof-track: look for parametric infinite families of solutions (to show $N(x)\gg x$) and then attempt to classify all solutions or at least show $N(x)=O(x)$. Disproof-track: search computationally for superlinear growth or irregular behavior.

WORK

FAST REALITY CHECK. A sieve computation of $\sigma(n)$ and a direct check of all $a+b\le x$ gives the exact counts:

x=1000:  N(x)=1620
x=2000:  N(x)=3806
x=5000:  N(x)=12484
x=10000: N(x)=30012
x=20000: N(x)=71612

In particular $N(x)/x$ is empirically about $1.62,1.90,2.50,3.00,3.58$ at these sample points. The smallest solutions include $(a,b,a+b)=(1,2,3)$ and $(2,6,8)$.

Lemma 1061.1 (proper-divisor reformulation). Let $s(n):=\sigma(n)-n$ be the sum of proper divisors of $n$ (with $s(1)=0$). Then \[ \sigma(a)+\sigma(b)=\sigma(a+b)\quad\Longleftrightarrow\quad s(a)+s(b)=s(a+b). \]

Proof. Using $\sigma(n)=n+s(n)$, \[ \sigma(a)+\sigma(b)= (a+s(a))+(b+s(b)) = (a+b) + (s(a)+s(b)). \] Also $\sigma(a+b)=(a+b)+s(a+b)$. Subtracting $(a+b)$ from both sides yields the equivalence. $\square$

Lemma 1061.2 (explicit infinite family). If $a\ge 1$ satisfies $\gcd(a,6)=1$, then $(a,2a)$ is a solution: \[ \sigma(a)+\sigma(2a)=\sigma(3a). \] Consequently $(2a,a)$ is also a solution.

Proof. Assume $\gcd(a,6)=1$, so in particular $\gcd(a,2)=\gcd(a,3)=1$. By multiplicativity of $\sigma$ on coprime inputs, \[ \sigma(2a)=\sigma(2)\sigma(a)=3\sigma(a),\qquad \sigma(3a)=\sigma(3)\sigma(a)=4\sigma(a). \] Therefore \[ \sigma(a)+\sigma(2a)=\sigma(a)+3\sigma(a)=4\sigma(a)=\sigma(3a), \] as required. $\square$

Corollary 1061.3 (linear lower bound). There exists an absolute constant $c_0>0$ such that $N(x)\ge c_0 x - O(1)$. In fact one may take $c_0=2/9$.

Proof. For each integer $a$ with $1\le a\le \lfloor x/3\rfloor$ and $\gcd(a,6)=1$, Lemma 1061.2 gives two ordered solutions $(a,2a)$ and $(2a,a)$ with sum $3a\le x$. The set of integers $a\le y$ coprime to $6$ has size $\ge \lfloor y/3\rfloor$ (one in each block of three consecutive integers avoids both factors $2$ and $3$), so the number of such $a$ with $a\le x/3$ is $\ge \lfloor x/9\rfloor$. Hence $N(x)\ge 2\lfloor x/9\rfloor = (2/9)x - O(1)$. $\square$

Lemma 1061.4 (no prime-sum solutions except $(1,2)$). If $a,b>1$ and $a+b$ is prime, then $\sigma(a)+\sigma(b)\ne \sigma(a+b)$. Equivalently, the only solutions with $a+b$ prime are $(a,b)=(1,2)$ and $(2,1)$.

Proof. If $n>1$ then $\sigma(n)\ge n+1$ because $1$ and $n$ are divisors. Thus for $a,b>1$, \[ \sigma(a)+\sigma(b)\ge (a+1)+(b+1)=a+b+2. \] If $a+b=p$ is prime, then $\sigma(a+b)=\sigma(p)=p+1=a+b+1$. Therefore $\sigma(a)+\sigma(b)\ge a+b+2> a+b+1=\sigma(a+b)$, so equality is impossible. If one of $a,b$ equals $1$, then the only way to have $a+b$ prime and satisfy the equation is the direct check $(1,2)$ (and symmetry). $\square$

VERIFICATION

Lemma 1061.2 uses only the standard multiplicativity of $\sigma$ for coprime arguments and the explicit values $\sigma(2)=3$, $\sigma(3)=4$; the coprimality condition $\gcd(a,6)=1$ is exactly what is needed. The corollary's counting bound $\#\{a\le y:\gcd(a,6)=1\}\ge \lfloor y/3\rfloor$ was checked by the simple observation that in each triple $(3t+1,3t+2,3t+3)$ at least one number is not divisible by $2$ or $3$. The numerical counts for $N(x)$ were computed by a direct sieve for $\sigma$ and then checking all pairs with $a+b\le x$.

FINAL

UNRESOLVED

1. (i) Strongest proved partial result. There are infinitely many solutions; in fact the explicit family $(a,2a)$ with $\gcd(a,6)=1$ gives the lower bound $N(x)\ge (2/9)x-O(1)$ (Lemma 1061.2 and Corollary 1061.3). Computations show $N(20000)=71612$ ordered solutions.
2. (ii) First gap (crisp). Prove a matching upper bound $N(x)=O(x)$ and, stronger, show that the limit $\lim_{x\to\infty} N(x)/x$ exists and is positive (or compute it).
3. (iii) Top 3 next moves. $1$ Classify all solutions with a fixed ratio $b/a$ or with $\gcd(a,b)=1$, using the multiplicativity of $\sigma$. $2$ Prove that for each fixed sum $n$ the number of solutions to $\sigma(a)+\sigma(n-a)=\sigma(n)$ is uniformly bounded, implying $N(x)=O(x)$ by summing over $n\le x$. $3$ Extend computations to larger $x$ and try to identify additional parametric families beyond $(a,2a)$ with $\gcd(a,6)=1$.
4. (iv) Minimal counterexample structure. A counterexample to $N(x)\sim cx$ would likely involve sums $n$ admitting an unusually large number of representations, i.e. many $a$ with $\sigma(a)+\sigma(n-a)=\sigma(n)$. Such $n$ might need to be highly composite (large $\sigma(n)$) to have room for many decompositions.