Consider the symmetric boolean function $$F(x_1,\dots,x_n)=1\iff\sum_{i=1}^nx_i\mbox{ is a square}.$$

It is implementable in $TC^0$.

Is there an $ACC^0$ implementation?

The reason I ask is there seems to be few tricks which can be deployed.

  1. Square is sum of consecutive odd numbers.

  2. Square is $0$ or $1$ mod $4$ (checkable by $\oplus$ gates).

  3. Squares are the set of the integers having odd number of divisors.

Few are available in https://proofwiki.org/wiki/Category:Square_Numbers.

But is there a finite number of tricks which capture the function and place it in $ACC^0$?


1 Answer 1


Suppose $x \in \{0,\dots,2n\}$. Then we have $((2n)^2 - n) + x$ is square if and only if $x=n$. This is easily seen to imply that the square function is complete for $\mathrm{TC}^0$ under $\mathrm{AC}^0$ Turing reductions.

  • $\begingroup$ Can you provide why the function is TC0 hard? $\endgroup$
    – User2021
    Mar 1, 2021 at 15:08
  • $\begingroup$ Ah it captures threshold or rather equality. $\endgroup$
    – User2021
    Mar 1, 2021 at 15:12

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