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Feb 9, 2021 at 6:16 history edited admissiblecycle CC BY-SA 4.0
Following the comments of Dodd and Wojowu, the condition $p \neq 2$ has been added, since this is clearly necessary.
Feb 9, 2021 at 6:04 vote accept admissiblecycle
Feb 8, 2021 at 16:38 answer added R.P. timeline score: 3
Feb 7, 2021 at 22:02 comment added Wojowu By Jacobi's four square theorem, the number of solutions in $\frac{1}{5^k}\mathbb Z$, equal to the number if integer solutions to $x^2+y^2+z^2+w^2=5^{2k}$, is equal to $8(1+5+\dots+5^{2k})$, so we definitely have many solutions. This doesn't tell us how they are distributed, but at least we can't have finiteness issues like with $2$.
Feb 7, 2021 at 20:58 answer added R.P. timeline score: 2
Feb 7, 2021 at 19:33 comment added markvs Try replacing $5$ with $2$.
Feb 7, 2021 at 13:35 history asked admissiblecycle CC BY-SA 4.0