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1 vote

Only trivial solution to a pair of constrained linear diophantine equations

The answer to the first question is negative. Let $A$ denote the set of weights $\{a_i\}$. Strengthen the first constraint to $0 \le x_i \le 2$. If we have two different subsets $S_1, S_2 \subset A$ ...
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3 votes

$x^3+x^2y^2+y^3=7$, and solvable families of Diophantine equations

Let me post an elementary answer to this question, which I have found recently. (a) Assume that $(x,y)$ is an integer solution. From symmetry, we may assume that $|y|\geq |x|$. If fact, cases $|y|=|x|$...
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2 votes

Rational solutions to $P(x,y)=0$ for $P$ reducible over ${\mathbb C}$

This also follows from Prop. 2.3.26(i) in Bjorn Poonen's Rational Points on Varieties, where it is stated that if for a finite type $k$-scheme $X$ the set of rational points $X(k)$ is Zariski dense, ...
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4 votes

Rational solutions to $P(x,y)=0$ for $P$ reducible over ${\mathbb C}$

The main idea of the proof already appears in what you've written, but here are some more details. Factor $P(x,y) = Q_1(x,y) \cdots Q_n(x,y)$ into irreducibles, where the factorization takes place ...
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