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Prove that for every $n \geq 6$ the equality $1/a_1^2 + ... 1/a_n^2 = 1$ has answer in $\mathbb{Z}$ (I mean it has an answer where all $a_i \in \mathbb{Z}$) (Repetition is allowed)

After some testing I found the answer $6,2,2,2,3,3$ for $n=6$ and $4,4,4,4,2,2,2$ for $n=7$. But I don't know how can I generalize this. Maybe I need to use induction but I don't know how to get from $n$ to $n+1 $ because obviously the answer completely change every time.

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    $\begingroup$ If you can find it for $n$, then you can find it for $n+3$. Just replace $1/a_1^2$ by $1/(2a_1)^2+1/(2a_1)^2+1/(2a_1)^2+1/(2a_1)^2$ $\endgroup$ – Julian Mejia May 8 at 16:54
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If it is true for $n$, then it is true for $n+3$, because $\frac{1}{a_n^2}=\frac{1}{4a_n^2}+\frac{1}{4a_n^2}+\frac{1}{4a_n^2}+\frac{1}{4a_n^2}$. So you only need to find examples for $n=6,7,8$

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