0xPARC's mersenne prime puzzle

2026-09-16

# Question

Are there any nontrivial solutions to the system of equations

$$ \begin{aligned} a + b + c + d &\equiv 0 \pmod{p} &&\text{(1)} \\ a^2 + b^2 + c^2 + d^2 &\equiv 0 \pmod{p} &&\text{(2)} \\ a^3 + b^3 + c^3 + d^3 &\equiv 0 \pmod{p} &&\text{(3)} \end{aligned} $$

for $p = 2^{127} - 1$?

# Solution

# Collapsing the system into a single product

From (1) we can write:

$$ d \equiv -(a + b + c) \pmod{p} \tag{4} $$

Substituting (4) into (3) gives:

$$ a^3 + b^3 + c^3 - (a + b + c)^3 \equiv 0 \pmod{p} \tag{5} $$

Now we have an identity with these cubes, and that is:

$$ (a+b+c)^3 - (a^3+b^3+c^3) = 3(a+b)(b+c)(a+c) \tag{6} $$

We can write (6) as a congruence, since any equation can be written as a congruent relation in the mods. Combining it with (5), we get the following implication:

$$ \begin{aligned} 3(a+b)(b+c)(a+c) &\equiv 0 \pmod{p} \\ \implies (a+b)(b+c)(a+c) &\equiv 0 \pmod{p} \end{aligned} \tag{7} $$

Dropping the $3$ is allowed because $p \neq 3$, so $3$ is invertible mod $p$.

Now, $p$ is prime, and a product can only vanish mod a prime if one of its factors does — that’s Euclid’s lemma. So one of the three factors in (7) must be $\equiv 0 \pmod{p}$. They are interchangeable under a relabelling of $a$, $b$ and $c$, so we lose nothing by assuming it’s $(a+b)$:

$$ \begin{aligned} a+b &\equiv 0 \pmod{p} \\ \implies a &\equiv -b \pmod{p} \end{aligned} \tag{8} $$

And from (1), we can say:

$$ d \equiv -c \pmod{p} \tag{9} $$

# Bringing in the squares

Substituting (8) and (9) into (2):

$$ \begin{aligned} a^2 + b^2 + c^2 + d^2 &\equiv 0 \pmod{p} \\ \implies 2a^2 + 2c^2 &\equiv 0 \pmod{p} \\ \implies a^2 &\equiv -c^2 \pmod{p} \end{aligned} \tag{10} $$

Halving is fine here, since $p$ is odd. Now suppose $c \not\equiv 0 \pmod{p}$. Then $c$ is invertible mod $p$, and we can divide (10) through by $c^2$:

$$ \left(\frac{a}{c}\right)^2 \equiv -1 \pmod{p} \tag{11} $$

So the whole question comes down to this: does $-1$ have a square root mod $p$?

# Why (11) cannot happen

The following statement is true and well known in modular math:

No prime $p$ of the form $2^n - 1$, with $n \geq 2$, can divide $x^2 + 1$

Why’s that? Here’s a quick derivation — we’ll do it for our $p$, but the same argument works for any $n \geq 2$. We start by assuming that $x^2+1$ is indeed divisible by $p$:

$$ x^2 \equiv -1 \pmod{p} \tag{12} $$

Raising both sides of (12) to the power of $\frac{p-1}{2}$:

$$ x^{p-1} \equiv (-1)^{\frac{p-1}{2}} \pmod{p} \tag{13} $$

The LHS of (13) is where Fermat’s little theorem comes in. It states that:

If $p$ is a prime number and $p$ does not divide $a$, then $a^{p-1}$ leaves a remainder of $1$ when divided by $p$.

In our case, $p$ doesn’t divide $x$, it really can’t, because if it did, then $p$ would divide $x^2$ and hence could not divide $x^2+1$. So the LHS of (13) is just $1$.

And due to $p$ being a Mersenne prime, we have:

$$ \begin{aligned} p &= 2^{127} - 1 \\ \implies p - 1 &= 2^{127} - 2 \\ \implies \frac{p - 1}{2} &= 2^{126} - 1 \end{aligned} \tag{14} $$

And $2^{126} - 1$ is an odd number, which means the RHS of (13) just stays $-1$. So (13) becomes:

$$ \begin{aligned} 1 &\equiv -1 \pmod{p} \\ \implies 2 &\equiv 0 \pmod{p} \end{aligned} \tag{15} $$

Or, $p$ divides $2$ exactly. But this is only possible when $p$ is either $1$ or $2$. Our $p$ is very much larger than $2$, so our assumption (12) is false.

# Wrapping up

So $-1$ has no square root mod $p$, and (11) is impossible. That kills the supposition we made to get there, leaving $c \equiv 0 \pmod{p}$.

Everything else follows immediately. From (10), $a^2 \equiv -c^2 \equiv 0 \pmod{p}$, and since $p$ is prime, $a \equiv 0 \pmod{p}$ as well. From (9), $d \equiv -c \equiv 0$, and from (8), $b \equiv -a \equiv 0$. So every solution has

$$ a \equiv b \equiv c \equiv d \equiv 0 \pmod{p} \tag{16} $$

that is, all four are multiples of $p$. These are exactly the trivial solutions!

Hence, we have shown that there cannot be any nontrivial solutions to (1), (2) and (3).