# Is computing a square root of a number and having more than 2 roots a reliable way to decide primality?

I was reading CLRS and it mentions that if $p$ is a prime of the form $4k+3$ and $a$ is a quadratic residue, then $a^{k+1}$ is a square root of $a$. One can also easily show that $a^{-k}$ is a square root of $a$ as well.

Can we use this to get a primality test, for numbers $N$ of the form $N=4k+3$?

After reviewing Miller-Rabin (and after asking the related question Is there any efficient algorithm for primality testing for numbers that are of the form $4k+3$ using the square root function? ) and thinking about it a little more I thought of an algorithm that seems promising. The intuition is that if $N=pq$ (or some other composite) there will be more than 2 square roots. Therefore, if we can find more than 2 square roots, we have evidence that $N$ is a composite. With that idea, here is the algorithm:

PrimalityTest$(N)$: (where $N = 4k + 3$)

1. Choose $x \in \mathbb{Z}_N$ and compute $a = x^2 \pmod N$.
2. Compute $y = a^{(N+1)/4} \pmod N$. (Notice that if $N$ is prime, this is just computing a square root of $a$.)
3. If $y^2 \equiv a \pmod N$ but $y \not \equiv \pm x \pmod N$ then there are more than 2 distinct square roots of $a$. Therefore, return "composite".
4. If $y^2 \not \equiv a \pmod N$ then return "composite". (If $N$ were prime, then step 2 would have computed a valid square root of $a$, so by the contrapositive, if $y$ is not a valid square root, then $N$ is not prime.)
5. If $y^2 \equiv a \pmod N$ and $y \equiv \pm x \pmod N$ we have no idea what $N$ is. Therefore, go back to step 1 and choose a different $x$. If this step happens too often, simply return "prime".

I believe this will succeed with probability 1/2, since it has half chance of choosing a "bad" square root that gives us no information. The only step I am not sure how to analyze probabilistically is step 4, but it seems that step 2 might do nonsense when $N$ is composite, it seems the probability that 4 catches is is quite high. I'd guess around $\frac{N-2}{N}$.

• I think your analysis of success probability needs to be fleshed out a bit more. 1. I can't understand what you are trying to say in the last sentence of this question. Can you edit to clarify/elaborate on that? 2. Can you define what you mean by a "bad" square root? – D.W. Dec 14 '15 at 20:56

Recall that $N$ is a Carmichael number if $x^{N-1}\equiv 1 \pmod N$ holds for all $x \in \mathbb{Z}_N^*$. All Carmichael numbers are odd. I'll define a super-Carmichael number to be a number $N$ such that $x^{(N-1)/2}\equiv 1 \pmod N$ holds for all $x \in \mathbb{Z}_N^*$.
Now consider what happens in your algorithm if $N$ is a super-Carmichael number. We pick $x$ randomly, then compute
$$y=a^{(N+1)/4}=(x^2)^{(N+1)/4}=x^{(N+1)/2} = x^{(N-1)/2} \cdot x = x \pmod N,$$
since $x^{(N-1)/2} = 1 \pmod N$. It follows that we also have $y^2 = x^2 \pmod N$. Thus, no matter what value of $x$ you choose, you'll always find that $y^2 \equiv a \pmod N$ is true and that $y \equiv \pm x \pmod N$ (when $N$ is a super-Carmichael number). So, your algorithm will always give an incorrect answer, when you run it on a super-Carmichael number.
For instance, $1729 = 7 \times 13 \times 19$ is a super-Carmichael number: we have $x^{36} \equiv 1 \pmod{1729}$ for all $x \in \mathbb{Z}_{1729}^*$, and $36$ divides $(1729-1)/2 = 864$. Therefore, your algorithm gives an incorrect answer on 1729. I suspect there are infinitely many super-Carmichael numbers. I also suspect that your algorithm might actually give an incorrect answer on all Carmichael numbers (not just the super-Carmichael ones), but I haven't tried to check the details to see if that's actually true.