For an odd prime $p$, we say $f(X) \in {\mathbb F}_p[X]$ computes square roots in $\mathbb F_p$ if, for all nonzero perfect squares $a \in \mathbb F_p$, we have $f(a)^2 = a$. When $p \equiv 3 \mod 4$, it is well known that $f(X) = X^{(p+1)/4}$ computes square roots. This degree is surprisingly low (and in fact lowest possible), since we have specified $(p-1)/2$ evaluations (up to sign) of the polynomial $f(X)$. On the other hand, for $p \equiv 1 \mod 4$ there was previously no nontrivial bound known on the lowest degree of a polynomial computing square roots in $\mathbb F_p$; it could have been anywhere between $\frac{p}{4}$ and $\frac{p}{2}$. We show that for all $p \equiv 1 \mod 4$, the degree of a polynomial computing square roots has degree at least $p/3$. Our main new ingredient is a general lemma which may be of independent interest: powers of a low degree polynomial cannot have too many consecutive zero coefficients. The proof method also yields a robust version: any polynomial that computes square roots for 99\% of the squares also has degree almost $p/3$. In the other direction, we also show that for infinitely many $p \equiv 1 \mod 4$, the degree of a polynomial computing square roots can be $(\frac{1}{2} - \Omega(1))p$.

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