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.

  • 1
    $\begingroup$ 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

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 $n=6,7,8$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.