Why is $120^{128}+1$ a prime number

Question

Wolfram Alpha says that the number $120^{128}+1$ is prime. I wonder if that is in fact true, and if so, what is an argument for it.

Thank you for your interest!

Answer 1

If given the correct command, WolframAlpha can tell us that $22$ is a primitive root modulo $120^{128}+1$.

• PowerMod[22, 120^128/{1, 2, 3, 5}, 120^128 + 1] computes the list $\{ 22^{120^{128}}, 22^{120^{128}/2}, 22^{120^{128}/3}, 22^{120^{128}/5} \}$ modulo $120^{128}+1$.
• The fact that $22^{120^{128}} \equiv 1 \pmod {120^{128}+1}$ tells us that the order of $22$ modulo $120^{128}+1$ is a divisor of $120^{128}$.
• The fact that $22^{120^{128}/2} \not\equiv 1 \pmod {120^{128}+1}$ tells us that the order of $22$ modulo $120^{128}+1$ is not a divisor of $120^{128}/2$. Similarly, the order of $22$ modulo $120^{128}+1$ is not a divisor of $120^{128}/3$ nor a divisor of $120^{128}/5$.
• Since $2$, $3$, and $5$ are the only primes dividing $120^{128}$, we conclude that the only divisor of $120^{128}$ that is not a divisor of any of $120^{128}/2$, $120^{128}/3$, or $120^{128}/5$ is $120^{128}$ itself.
• Therefore the order of $22$ modulo $120^{128}+1$ is exactly $120^{128}$, which implies that $120^{128}+1$ is prime.

($22$ is simply the first primitive root I found; the smallest three primitive roots are $19$, $22$, and $29$.)

Answer 2

I can't find any public documentation of how WolframAlpha does primality testing. The simplest generally applicable option for numbers this large is the Fermat primality test, which for a positive integer $p$ means randomly choosing integers $a \in [2, p-2]$ and testing whether or not $a^{p-1} \equiv 1 \bmod p$. With $p = 120^{168}$ this computation can be done using binary exponentiation without too much hassle; the only prime factors of $120$ are $2, 3, 5$ we could do it by repeatedly taking second, third, and fifth powers, and for fifth powers we could use $5 = 2^2 + 1$ to speed the calculation up a bit too.

Unfortunately the Fermat primality test is really a compositeness test; if it finds $a$ such that $a^{p-1} \not \equiv 1 \bmod p$ then $p$ is composite, but if it doesn't then $p$ is only a probable prime. By finding more and more values of $a$ that probability can be driven arbitrarily low but it's still not a proof. There are variants here like the Solovay-Strassen primality test, the Miller-Rabin primality test, or the Baillie-PSW primality test.

For $p$ such that the prime factorization of $p - 1$ is known we can also apply the test Greg Martin describes which is called the Lucas primality test, which I just learned about right now. This one really is a primality test: if a number passes then it is provably prime.

Mathematica has a function ProvablePrimeQ which generates certificates proving that numbers are prime based on elliptic curve primality tests; I don't know if this is what WolframAlpha is doing. It might be some heuristic combination of tests depending on what seems like it would be easiest to apply?

Edit: Documentation linked to on Wikipedia says that Mathematica's PrimeQ function uses a variation of the Baillie-PSW test:

PrimeQ first tests for divisibility using small primes, then uses the Miller–Rabin strong pseudoprime test base 2 and base 3, and then uses a Lucas test. There are no known composite numbers that pass this procedure.