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Consider the ring $\mathbb{Z}_3$ of 3-adic integers. Does there exist a positive integer $n$ and a solution to $(X_1^2 + X_2^2 + \cdots + X_{n - 1}^2)^2 = 2X_n^4$ in $\mathbb{Z}_3^n$? If so, what is the smallest $n$ for which a solution exists?

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Won't [p-adic-number-theory] fit here even better? –  Asaf Karagila Feb 23 '12 at 15:23
    
@Asaf: sure. Please feel free to change it... –  Pete L. Clark Feb 23 '12 at 15:24
    
@Pete: Thanks for the confirmation. Since the p-adic tag has so little questions I just added it as the exposure of this question would decrease dramatically (183 followers vs. 7 followers...) –  Asaf Karagila Feb 23 '12 at 15:34

1 Answer 1

up vote 6 down vote accepted

There are no solutions with $X_n \neq 0$, for then dividing through by $X_n^4$ would yield that $2$ is a square in $\mathbb{Q}_3$...which it isn't.

Solutions with $X_n = 0$ correspond to tuples $(X_1,\ldots,X_{n-1})$ such that $X_1^2 + \ldots + X_{n-1}^2 = 0$, i.e., we want the sum of $n-1$-squares to be an isotropic quadratic form. This occurs over $\mathbb{Z}_3$ (equivalently, over $\mathbb{Q}_3$) iff $n-1 \geq 3$.

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Oh, I see. I guess your last statement is because of Hensel's lemma. –  Evariste Feb 23 '12 at 15:32
    
@Evariste: yes, that's right. –  Pete L. Clark Feb 23 '12 at 17:23

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