We find the complete integer solutions of the equation X2+Y2+Z2-4⁢X⁢Y-4⁢Y⁢Z+10⁢X⁢Z=1X^{2}+Y^{2}+Z^{2}-4XY-4YZ+10XZ=1. As an application, we prove that, for each solution (a,b,c)(a,b,c) such that a>0a>0, b-2⁢a>0b-2a>0 and (b-2⁢a)2≥4⁢a(b-2a)^{2}\geq 4a, there is a vector bundle 𝐸 on P3\mathbb{P}^{3} defined by a minimal linear resolution 0→OP3⁢(-2)a→OP3⁢(-1)b→OP3c→E→00\to\mathcal{O}_{\mathbb{P}^{3}}(-2)^{a}\to\mathcal{O}_{\mathbb{P}^{3}}(-1)^{b}\to\mathcal{O}_{\mathbb{P}^{3}}^{c}\to E\to 0. In particular, 𝐸 satisfies χ⁢(End⁡E)=1\chi(\operatorname{End}E)=1.

中文翻译：

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ℙ3上具有同维2且𝜒（End𝐸）= 1的向量束𝐸

我们找到方程X2 + Y2 +Z2-4⁢X⁢Y-4⁢Y⁢Z+10⁢X⁢Z= 1X ^ {2} + Y ^ {2} + Z ^ {2}的完整整数解-4XY-4YZ + 10XZ = 1。作为应用，我们证明对于每个解（a，b，c）（a，b，c）使得a> 0a> 0，b-2 ba> 0b-2a> 0和（b-2⁢ a）2≥4⁢a（b-2a）^ {2} \ geq 4a，在P3 \ mathbb {P} ^ {3}上有一个向量束𝐸，其定义为最小线性分辨率0→OP3⁢（-2 ）a→OP3⁢（-1）b→OP3c→E→00 \ to \ mathcal {O} _ {\ mathbb {P} ^ {3}}（-2）^ {a} \ to \ mathcal {O} _ {\ mathbb {P} ^ {3}}（-1）^ {b} \至\ mathcal {O} _ {\ mathbb {P} ^ {3}} ^ {c} \至E \至0。特别地，𝐸满足χ⁢（End⁡E）＝ 1 \ chi（\运算符{End} E）＝ 1。