GPT 6 Astra Ultra attempt on Math-5585
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Erdős Problem 334: a sum of two smooth numbers
Statement and scope
Put $P^+(1)=1$ and let $P^+(n)$ be the largest prime divisor of $n>1$. Define \[ g(n)=\min_{1\le a<n}\max(P^+(a),P^+(n-a)). \] The question asks for the best uniform growth bound on $g$. We prove the elementary bounds \[ g(n)\le\lfloor\sqrt n\rfloor+1,\qquad g(n)>(\log n)^{2-\epsilon}\quad\hbox{for a density-one set of }n \tag{1} \] for each fixed $0<\epsilon<2$. The optimal asymptotic bound is unresolved here.These replace the earlier GPT-5.2 attempt's essentially tautological estimates. They do not improve the known uniform bound: https://www.erdosproblems.com/334 records Balog's stronger exponent $4/(9\sqrt e)+\epsilon$. That result is context and is not used below. The proof consists of elementary factorization and the standard Rankin counting method.
An explicit square-root decomposition
For $n\ge4$, put $m=\lfloor\sqrt n\rfloor\ge2$ and write $n=mq+r$, $0\le r<m$. Then $m\le q\le m+2$. If $r>0$, use \[ n=m(q-r)+(m+1)r. \tag{2} \] Both summands are positive: $q-r\ge m-(m-1)=1$. Also $q-r\le q-1\le m+1$, so every factor displayed in $2$ is at most $m+1$. Each summand is therefore $(m+1)$-smooth.If $r=0$, use $n=m+m(q-1)$. Here $q-1\ge1$ and $q-1\le m+1$, giving the same bound. The cases $n=2,3$ follow from $1+1$ and $1+2$.
A counting obstruction near the logarithmic-square scale
Let $\Psi(X,y)$ count the positive $y$-smooth integers at most $X$. For $0<\sigma<1$, positivity and unique factorization give \[ \Psi(X,y) \le X^\sigma\sum_{P^+(a)\le y}a^{-\sigma} =X^\sigma\prod_{p\le y}(1-p^{-\sigma})^{-1}. \] Using $-\log(1-t)\le t/(1-t)$ and summing over all integers instead of just primes, \[ \log\prod_{p\le y}(1-p^{-\sigma})^{-1} \le\frac1{1-2^{-\sigma}}\sum_{p\le y}p^{-\sigma} \ll_\sigma y^{1-\sigma}. \tag{3} \]Fix $0<\epsilon<2$, take $y=(\log X)^{2-\epsilon}$ and $\sigma=1/2-\epsilon/8>0$. Since \[ (2-\epsilon)(1-\sigma) =(2-\epsilon)(1/2+\epsilon/8) =1-\epsilon/4-\epsilon^2/8<1, \] equation $3$ is $o(\log X)$. Consequently \[ \Psi(X,y)\le X^{1/2-\epsilon/8+o(1)}. \] There are at most $\Psi(X,y)^2$ sums of two such smooth integers, so \[ \#\{n\le X:g(n)\le(\log X)^{2-\epsilon}\} \le X^{1-\epsilon/4+o(1)}=o(X). \tag{4} \] For all sufficiently large $n\le X$, $(\log n)^{2-\epsilon}\le(\log X)^{2-\epsilon}$. Thus $4$ proves the density-one lower assertion in $1$. In particular no fixed smoothness threshold can work for every large integer.