7

At Twelve new primitive binary trinomials, $x^{74207281}+x^{9999621}+1$ is shown to be a primitive trinomial in $GF(2)[x]$.

Note that $2^{74207281}-1$ was the largest known (Mersenne) prime before 2018, that is the prime in the exponent of the Mersenne prime gave rise to a trinomial.

In the interim, new Mersenne primes have been found.

Just announced (Dec 2018), $2^{82589933}-1$ is prime, and in Jan 2018, $2^{77232917}-1$ was found to be prime.

Do these also give rise to trinomials?

To be explicit, are there $j$ or $k$ such that $x^{82589933}+x^{j}+1$ or $x^{77232917}+x^{k}+1$ are primitive trinomials in $GF(2)[x]$?

quid
  • 42,135
Ed Pegg
  • 20,955
  • 2
    I don't understand the votes to put this on hold. There was nothing unclear about the question. Anyway, when $p=2^m-1$ is a Mersenne prime, a polynomial $f(x)$ of degree $m$ is primitive in $\Bbb{F}_2[x]$ if and only if it is irreducible if and only if $x^{2^m}\equiv x\pmod{f(x)}$. That last congruence can be checked with the good ole square-and-multipl if you can spare the CPU time. – Jyrki Lahtonen Dec 23 '18 at 04:04
  • 2
    Having said that, I'm not entirely sure that people here are well placed to answer this. I would guess that the search for such trinomials does take quite a bit of CPU resources even though testing any candidate is relatively straightforward. – Jyrki Lahtonen Dec 23 '18 at 04:13
  • 1
    I want to make it clear that the method I used here to find primitive trinomials of degree $127$ absolutely depended on the coincidence that $127$ is itself also a Mersenne prime. That trick doesn't work here. – Jyrki Lahtonen Dec 27 '18 at 17:09
  • 1
    I rearranged the question a bit; even the edited version, I found it hard to follow. – quid Dec 28 '18 at 13:49

1 Answers1

4

As per Swan's Theorem, reproduced in The Great Trinomial Hunt, $T_{r,s}(x) \equiv x^r + x^s + 1$ has an even number of factors over $GF(2)$ if $r$ is $\pm 3 \pmod{8}$ and $s$ is even and not a factor of $2r$. Given that both of the Mersenne exponents above (82589933 and 77232917) are 5 mod 8, if $s>2$ even then we know $T_{r,s}$ is reducible.

Likewise as stated in the paper, $x^r + x^s + 1$ has the same number of factors as $x^r + x^{r-s} + 1$, which eliminates all odd $s$ aside from $s = r-2$.

So it suffices to look at $x^r + x^2 + 1$ for the two values of $r$ in question. This is not exactly an easy computation, even though polynomial squaring is easy in $GF(2)$.

Fortunately, the paper then suggests a useful trick -- sieving against small factors by using the fact that $T \mod x^d+1 = x^{(r \pmod{d})} + x^{(s\pmod{d})} + 1$.

With this, it is easy to see that $x^{82589933} + x^2 + 1$ and $x^{77232917} + x^2 + 1$ are both divisible by $x^3+x+1$.

So there are no primitive trinomials for these primes.

Craig
  • 3,536
  • Wow! I need to find the time to read that article! The claim in the next to last paragraph is paper and pencil work: $x^3+x+1$ is a factor of $x^7+1$, and both the large exponents are $\equiv6\pmod7$. $x^6+x^2+1=(x^3+x+1)^2$. – Jyrki Lahtonen Jan 06 '21 at 22:57