# Prove that for every $n\geq6$ the equation $1/a_1^2 + … 1/a_n^2 = 1$ has answer in $\mathbb{Z}$

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.

• If you can find it for $n$, then you can find it for $n+3$. Just replace $1/a_1^2$ by $1/(2a_1)^2+1/(2a_1)^2+1/(2a_1)^2+1/(2a_1)^2$ – Julian Mejia May 8 at 16:54

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 $$n=6,7,8$$