GPT 6 Astra Ultra attempt on Math-5677
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Erdős Problem #476: the restricted two-fold sumset
Complete polynomial-method reconstruction, 24 September 2026.Statement
For $A\subseteq\mathbb F_p$, put $A\hat+ A=\{a+b:a,b\in A,\ a\ne b\}$. We prove \[|A\hat+ A|\ge\min(p,2|A|-3).\] The original theorem is due to Dias da Silva and Hamidoune. This proof reconstructs the elementary method of Alon, Nathanson and Ruzsa, including the coefficient identity it needs.A coefficient identity on a rectangular grid
Let $U,V$ be subsets of a field, with sizes $s+1,t+1$, and let $F(X,Y)$ have total degree at most $s+t$. Then \[[X^sY^t]F=\sum_{u\in U,v\in V}\frac{F(u,v)}{\prod_{u'\in U\setminus\{u\}}(u-u')\prod_{v'\in V\setminus\{v\}}(v-v')}.\tag{1}\] All denominators are nonzero. To prove $1$, one-variable Lagrange interpolation shows that the functional \[L_U(X^j)=\sum_{u\in U}\frac{u^j}{\prod_{u'\ne u}(u-u')}\] is zero for $j<s$ and one for $j=s$. For a monomial $X^iY^j$ with $i+j\le s+t$, either $(i,j)=(s,t)$ or at least one exponent is below its respective threshold. Applying $L_U L_V$ proves $1$ by linearity. In particular, a polynomial vanishing on $U\times V$ has zero coefficient on the left.The bound
Write $n=|A|$. The cases $n\le1$ are immediate, and $p=2$ is checked directly. Suppose henceforth $p$ is odd and $n\ge2$.If $2n-3\ge p$, then for any $c\in\mathbb F_p$, \[|A\cap(c-A)|\ge2n-p\ge3.\] At most one of these elements satisfies $a=c-a$. Any other gives distinct elements $a,c-a\in A$, so every $c$ belongs to the restricted sumset.
Now suppose $2n-3<p$ and, for contradiction, $|A\hat+ A|\le2n-4$. Enlarge this sumset to a set $C\subseteq\mathbb F_p$ of size $2n-4$, and define \[F(X,Y)=(X-Y)\prod_{c\in C}(X+Y-c).\] This polynomial vanishes on $A\times A$: off the diagonal one product factor vanishes, and on the diagonal $X-Y$ vanishes. Choose any $(n-1)$-element subset $B\subset A$ and use $1$ on $A\times B$, with $(s,t)=(n-1,n-2)$. Since $\deg F=2n-3=s+t$, its indicated coefficient must be zero. But that coefficient is \[\binom{2n-4}{n-2}-\binom{2n-4}{n-1} =\frac{(2n-4)!}{(n-1)!(n-2)!},\tag{2}\] where the second binomial is interpreted as zero when $n=2$. Every factorial argument in $2$ is less than $p$, so $2$ is nonzero in $\mathbb F_p$. This contradiction completes the proof.
Sharpness and scope
For $A=\{0,1,\ldots,n-1\}$ with $2n-3\le p$, the distinct-element sums are precisely the residues of $1,2,\ldots,2n-3$, giving equality. Large sets give the other sharp regime, where every residue occurs. The broader many-summand statement mentioned in the editorial is not needed for this two-summand problem and is not claimed proved here.SOLVED. Source statement: https://www.erdosproblems.com/476. Primary proof source: N. Alon, M. B. Nathanson and I. Z. Ruzsa, Adding distinct congruence classes modulo a prime, https://www.math.tau.ac.il/~nogaa/PDFS/annr3.pdf. The known argument is reproduced with attribution; the page's Lean label is not a claim that this text was formally checked.