Python 2, 56 bytes

n**(n*n-n)/(((2**n**n+1)**n**n>>n**n*~-n)%2**n**n)%n>n-2

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This is a proof-of-concept that this challenge is doable with only arithmetic operators, in particular without bitwise |, &, or ^. The code uses bitwise and comparison operators just for golfing, and they can easily be replaced with arithmetic equivalents.

However, the solution is extremely slow, and I haven't been able to run \$n=6\$, thanks to two-level exponents like \$2^{n^n}\$.

The main idea is to make an expression for the factorial \$n!\$, which lets us do a Wilson's Theorem primality test \$(n-1)! \mathbin{\%} n > n-2 \$ where \$ \mathbin{\%}\$ is the modulo operator.

We can make an expression for the binomial coefficient, which is made of factorials

$$\binom{m}{n} \ = \frac{m!}{n!(m-n)!}$$

But it's not clear how to extract just one of these factorials. The trick is to hammer apart \$n!\$ by making \$m\$ really huge.

$$\binom{m}{n} \ = \frac{m(m-1)\cdots(m-n+1)}{n!}= \frac{m^n}{n!}\cdot \left(1-\frac{1}{m}\right)\left(1-\frac{2}{m}\right)\cdots \left(1-\frac{n-1}{m}\right)$$

So, if we let \$c \$ be the product \$ \left(1-\frac{1}{m}\right)\left(1-\frac{2}{m}\right)\cdots \left(1-\frac{n-1}{m}\right)\$, we have

$$n! = \frac{m^n}{\binom{m}{n}} \cdot c$$

If we could just ignore \$c\$, we'd be done. The rest of this post is looking how large we need to make \$m\$ for this to work.

Note that \$c\$ approaches \$1\$ from below as \$ m \to \infty \$. We just need to make \$m\$ huge enough that omitting \$c\$ gives us a value with integer part \$n!\$ so that we may compute

$$n! = \left\lfloor \frac{m^n}{\binom{m}{n}} \right\rfloor $$

For this, it suffices to have \$1 - c < 1/n!\$ to avoid the ratio passing the next integer \$n!+1\$.

Observe that \$c\$ is a product of \$n\$ terms of which the smallest is \$ \left(1-\frac{n-1}{m}\right)\$. So, we have

$$c > \left(1-\frac{n-1}{m}\right)^n > 1 - \frac{n-1}{m} n > 1-\frac{n^2}{m},$$

which means \$1 - c < \frac{n^2}{m}\$. Since we're looking to have \$1 - c < 1/n!\$, it suffices to take \$m \geq n! \cdot n^2\$.

In the code, we use \$m=n^n\$. Since Wilson's Theorem uses \$(n-1)!\$, we actually only need \$m \geq (n-1)! \cdot (n-1)^2\$. It's easy to see that \$m=n^n\$ satisfies the bound for small \$n\$ and quickly outgrows the right hand side asymptotically, say with Stirling's approximation.

This answer doesn't use any number-theoretic cleverness. It spams Python's bitwise operators to create a manual "for loop", checking all pairs \$1 \leq i,j < n\$ to see whether \$i \times j = n\$.

Python 2, way too many bytes (278 thanks to Jo King in the comments!)

((((((2**(n*n)/(2**n-1)**2)*(2**((n**2)*n)/(2**(n**2)-1)**2))^((n*((2**(n*n-n)/(2**n-1))*(2**((n**2)*(n-1))/(2**n**2-1))))))-((2**(n*n-n)/(2**n-1))*(2**((n**2)*(n-1))/(2**(n**2)-1))))&(((2**(n*(n-1))/(2**n-1))*(2**((n**2)*(n-1))/(2**(n**2)-1)))*(2**(n-1)))==0))|((1<n<6)&(n!=4))

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This is a lot more bytes than the other answers, so I'm leaving it ungolfed for now. The code snippet below contains functions and variable assignment for clarity, but substitution turns isPrime(n) into a single Python expression.

def count(k, spacing):
    return 2**(spacing*(k+1))/(2**spacing - 1)**2
def ones(k, spacing):
    return 2**(spacing*k)/(2**spacing - 1)

def isPrime(n):
    x = count(n-1, n)
    y = count(n-1, n**2)
    onebits = ones(n-1, n) * ones(n-1, n**2)
    comparison = n*onebits
    difference = (x*y) ^ (comparison)
    differenceMinusOne = difference - onebits
    checkbits = onebits*(2**(n-1))
    return (differenceMinusOne & checkbits == 0 and n>1)or 1<n<6 and n!=4

Why does it work?

I'll do the same algorithm here in base 10 instead of binary. Look at this neat fraction:

$$ \frac{1.0}{999^2} = 1.002003004005\dots $$

If we put a large power of 10 in the numerator and use Python's floor division, this gives an enumeration of numbers. For example, \$ 10^{15}/(999^2) = 1002003004 \$ with floor division, enumerating the numbers \$ 1,2,3,4 \$.

Let's say we multiply two numbers like this, with different spacings of zeroes. I'll place commas suggestively in the product.

$$ 1002003004 \times 1000000000002000000000003000000000004 = $$ $$ 1002003004,002004006008,003006009012,004008012016 $$

The product enumerates, in three-digit sequences, the multiplication table up to 4 times 4. If we want to check whether the number 5 is prime, we just have to check whether \$ 005 \$ appears anywhere in that product.

To do that, we XOR the above product by the number \$ 005005005\dots005 \$, and then subtract the number \$ 001001001\dots001 \$. Call the result \$d\$. If \$ 005 \$ appeared in the multiplication table enumeration, it will cause the subtraction to carry over and put \$ 999 \$ in the corresponding place in \$d\$.

To test for this overflow, we compute an AND of \$d\$ and the number \$ 900900900\dots900 \$. The result is zero if and only if 5 is prime.

43 bytes

(4**n+1)**n%4**n**2/n&2**(2*n*n+n)/-~2**n<1

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The method is similar to Dennis' second (deleted) answer, but this answer is easier to be proved correct.

Proof

Short form

The most significant digit of (4**n+1)**n%4**n**2 in base \$2^n\$ that is not divisible by \$n\$ will make the next (less significant) digit in (4**n+1)**n%4**n**2/n nonzero (if that "next digit" is not in the fractional part), then a & with the bitmask 2**(2*n*n+n)/-~2**n is executed to check if any digit at odd position is nonzero.

Long form

Let \$[a_n,\dots,a_1,a_0]_b\$ be the number having that base \$b\$ representation, i.e., \$a_nb^n+\dots+a_1b^1+a_0b^0\$, and \$a_i\$ be the digit at "position" \$i\$ in base \$b\$ representation.

• \$\texttt{2**(2*n*n+n)/-~2**n} =\lfloor{2^{(2n+1)n}\over1+2^n}\rfloor =\lfloor{4^{n^2}\times 2^n\over1+2^n}\rfloor =\lfloor{{(4^{n^2}-1)\times 2^n\over1+2^n} +{2^n\over1+2^n}}\rfloor \$.

Because \$2^n\times{4^{n^2}-1\over1+2^n} =2^n(2^n-1)\times{(4^n)^n-1\over4^n-1} =[2^n-1,0,2^n-1,0,2^n-1,0]_{2^n}\$ (with \$n\$ \$2^n-1\$s) is an integer, and \$\lfloor{2^n\over1+2^n}\rfloor=0\$, 2**(2*n*n+n)/-~2**n = \$[2^n-1,0,2^n-1,0,2^n-1,0]_{2^n}\$.

Next, consider $$\begin{align} \texttt{(4**n+1)**n} &=(4^n+1)^n \\ &=\binom n04^{0n}+\binom n14^{1n}+\dots+\binom nn4^{n^2} \\ &=\left[\binom nn,0,\dots,0,\binom n1,0,\binom n0\right]_{2^n} \end{align}$$

\$4^{n^2}=(2^n)^{2n}\$, so %4**n**2 will truncate the number to \$2n\$ last digits - that excludes the \$\binom nn\$ (which is 1) but include all other binomial coefficients.

About /n:

• If \$n\$ is a prime, the result will be \$\left[\binom n{n-1}/n,0,\dots,0,\binom n1/n,0,0\right]_{2^n}\$. All digits at odd position are zero.

• If \$n\$ is not a prime:

  Let \$a\$ be the largest integer such that \$n\nmid\binom na\$ (\$n>a>0\$). Rewrite the dividend as

  \$\left[\binom n{n-1},0,\binom n{n-2},0,\dots,\binom n{a+1}, 0,0,0,\dots,0,0,0\right]_{2^n} + \left[\binom na,0,\binom n{a-1},0,\dots,\binom n0\right]_{2^n}\$

  The first summand has all digits divisible by \$n\$, and the digit at position \$2a-1\$ zero.

  The second summand has its most significant digit (at position \$2a\$) not divisible by \$n\$ and (the base) \$2^n>n\$, so the quotient when dividing that by \$n\$ would have the digit at position \$2a-1\$ nonzero.

  Therefore, the final result ((4**n+1)**n%4**n**2/n) should have the digit (base \$2^n\$, of course) at position \$2a+1\$ nonzero.

Finally, the bitwise AND (&) performs a vectorized bitwise AND on the digits in base \$2^n\$ (because the base is a power of 2), and because \$a\texttt &0=0,a\texttt&(2^n-1)=a\$ for all \$0\le a<2^n\$, (4**n+1)**n%4**n**2/n&2**(2*n*n+n)/-~2**n is zero iff (4**n+1)**n%4**n**2/n has all digits in first \$n\$ odd positions zero - which is equivalent to \$n\$ being prime.