Fermat Prime Puns

Link here. Quite short and accessible. Also contains other neat & similarly inclined proofs towards other results. Thought it was worth sharing!

I was wondering a while back if any finite fields could be considered as analogous to the Gaussian (complex) integers, rather than the usual real integers. Arithmetic on finite fields of prime order can all be modeled as arithmetic on the integers modulo that prime. Could something similar be done with Gaussian integers?

As it turns out the answer is yes, but only for a few exceptional cases. For a finite field to behave like the Gaussian integers it must contain an element that behaves analogously to the imaginary unit i, that is, its square must be congruent to -1. The most obvious cases are then the finite fields whose order is a prime that is one greater than a square number. The only such primes are all Fermat primes, except for 2 which is a trivial case.

GF(65537) is the finite field whose order is the largest Fermat prime. Instead of the usual way of specifying its elements (by the integers modulo 65,537 they are congruent to), we can instead introduce an imaginary unit i, which is congruent to 256 (so that i^2 is congruent to -1). The other imaginary unit -i is then congruent to 65,281.

Each element in GF(65537) could then be represented as a sum of an real and imaginary integer, treating 256 - i and 256i + 1 as congruent to zero. These congruence relations function analogously to moduli, so that each element of GF(65537) is mapped to a congruence class of Gaussian integers that forms a slightly rotated square grid of side length sqrt(65537). The finite fields GF(257), GF(17) and GF(5) all behave similarly but GF(3), derived from smallest Fermat prime, does not (since 2 is not a square number). So we have four finite fields that behave like Gaussian integers.

Not sure if this has any practical application but it sure is interesting to me.

I know that if p is prime and a is not divisible by p then this congruency holds.

But I'm not able to understand how and why exactly the algorithm works. Can someone please explain it in a simple way? How would I check whether a given number n, where 1 <= n <= 10^(12) is prime?

I think I have a proof for Fermat's Theorem for all prime exponents that is only two pages. Has this already been done?

So Fermat had a theorem about the sum of two squares equaling a prime number if said prime number has a remainder of 1 when divided by 4. And it seems to be true for numbers other than prime numbers.

Well, it made me think a bit about something.

Since it is known whether or not a number can be expressed as the sum of two squares just by dividing by 4, is there something like that, for say, the sum of two cubes? What about the sum of three cubes? Or, more generally, something like this:

Is there a way to check whether a number can be expressed as k numbers each raised to the nth power? And what about k = n?

Hi there,

I am totally lost on how to prove the following statement:

If n is not a power of 2, then 2^n + 1 is not a prime. I tried playing around with factoring something of the form x^m+y^m but realized quickly that's easier said than done.

i.e. infinitely many primes of the form 2^n + 1

Currently the only Fermat primes known are 3, 5, 17, 257, 65537.

After explaining Fermat's little theorem, Recreations in the Theory of Numbers says: "In Mersenne's numbers, that is, numbers of the form 2^n - 1, the exponent n is always a prime and therefore odd (except for n = 2, a trivial case). Hence, since in Fermat's theorem p is also a prime and odd, it follows that mn must be even, and therefore m must be even, equal to, say 2r, and then p = 2rn+1. [Here is where I get confused] This tells us, therefore, that if, for example, 2^11 -1 has a prime divisor, p (if it has any divisor, it has a prime divisor, of course), it must be of thr form 2r·11+1 = 22r+1, and in fact when r = 1, the prime 23 divides 2^11 - 1."

I feel like I understand Fermat's little theorem, but I don't see how the conclusion follows from it. Why does Fermat's little theorem imply all prime divisors of 2^n - 1 (when n is prime) must be of the form 2rn+1?

we begin with a simple lemma:

lemma: let f and g be functions, then f and g are linear and commute with each other
proof: every high school student knows this. ∎

theorem 1: every positive integer is a prime power
proof: use log(x+y)=log(x)+log(y)=log(xy) by distributing the log and recombining. taking exp of both sides gives x+y = xy for all x,y, so addition and multiplication are the same thing. it follows that fields and groups are the same thing. every finite field has prime power order, but there is a cyclic group of order n for all positive integers n, hence every positive integer is a prime power. ∎

theorem 2: fermats last theorem is false
proof: take the nth root of both sides of x^(n)+y^(n) = z^(n), swap the addition and the root using the lemma, and get x+y=z. choose your favourite integers that satisfy this equation, and you have a counterexample. ∎

theorem 3: the riemann hypothesis is true
proof: we have zeta(s) = 1^-s + 2^-s + 3^-s + ... = (1+2+3+...)^-s = (-1/12)^(-s), which is never zero. hence there are no nontrivial zeros, so the riemann hypothesis is vacuously true. ∎

FLINT headers:

#include "flint/flint.h" #include "flint/ulong_extras.h"

function:

static bool FermatProbablePrimalityTestFast(const mpz_class& n, unsigned int& nLength, CPrimalityTestParams& testParams, bool fFastFail = false) {

  mp_limb_t d = mpz_get_ui(n.get_mpz_t());

  if (n_is_probabprime_BPSW(d))
    return true;


  //if (mpz_probab_prime_p (n.get_mpz_t(), 12))
  //   return true;


// Faster GMP version
mpz_t& mpzE = testParams.mpzE;
mpz_t& mpzR = testParams.mpzR;

mpz_sub_ui(mpzE, n.get_mpz_t(), 1);
mpz_powm(mpzR, mpzTwo.get_mpz_t(), mpzE, n.get_mpz_t());
if (mpz_cmp_ui(mpzR, 1))
    return true;
if (fFastFail)
    return false;
// Failed Fermat test, calculate fractional length
mpz_sub(mpzE, n.get_mpz_t(), mpzR);
mpz_mul_2exp(mpzR, mpzE, nFractionalBits);
mpz_tdiv_q(mpzE, mpzR, n.get_mpz_t());
unsigned int nFractionalLength = mpz_get_ui(mpzE);
if (nFractionalLength >= (1 << nFractionalBits))
    return error("FermatProbablePrimalityTest() : fractional assert");
nLength = (nLength & TARGET_LENGTH_MASK) | nFractionalLength;
return false;

}

and

static bool FermatProbablePrimalityTest(const CBigNum& n, unsigned int& nLength) { CAutoBN_CTX pctx; CBigNum a = 2; // base; Fermat witness CBigNum e = n - 1; CBigNum r;

if (n_is_probabprime_BPSW(n.getulong()))
    return true;

BN_mod_exp(&r, &a, &e, &n, pctx);
if (r == 1)
    return true;

// Failed Fermat test, calculate fractional length
unsigned int nFractionalLength = (((n-r) << nFractionalBits) / n).getuint();
if (nFractionalLength >= (1 << nFractionalBits))
    return error("FermatProbablePrimalityTest() : fractional assert");
nLength = (nLength & TARGET_LENGTH_MASK) | nFractionalLength;
return false;

}

I'm writing a book on number theory, so naturally I'm wanting to give the reader a nice example of a nice result from number theory. One way to do this would be to provide an explanation of how a particular kind of integral was proved. But I'm not sure how to go about doing this in a way that's understandable to the layman. For example, I can list a bunch of examples of integrals, but I don't have any experience with them.

The best I can think of doing is to list one integral, then give a hint about how to do it, but then give an example of a particularly difficult case of the integral. Does anyone have any advice on how to approach this problem?

all these dooty stowey-memers is make me so hard, good sir

thank mr skeltal

I don't want to step on anybody's toes here, but the amount of non-dad jokes here in this subreddit really annoys me. First of all, dad jokes CAN be NSFW, it clearly says so in the sub rules. Secondly, it doesn't automatically make it a dad joke if it's from a conversation between you and your child. Most importantly, the jokes that your CHILDREN tell YOU are not dad jokes. The point of a dad joke is that it's so cheesy only a dad who's trying to be funny would make such a joke. That's it. They are stupid plays on words, lame puns and so on. There has to be a clever pun or wordplay for it to be considered a dad joke.

Again, to all the fellow dads, I apologise if I'm sounding too harsh. But I just needed to get it off my chest.

Do your worst!

I'm surprised it hasn't decade.

For context I'm a Refuse Driver (Garbage man) & today I was on food waste. After I'd tipped I was checking the wagon for any defects when I spotted a lone pea balanced on the lifts.

I said "hey look, an escaPEA"

No one near me but it didn't half make me laugh for a good hour or so!

Edit: I can't believe how much this has blown up. Thank you everyone I've had a blast reading through the replies 😂

It really does, I swear!

Because she wanted to see the task manager.

Hello,

here's a short question from elementary number theory / modular exponentiation.

Let p > 2 be prime, and consider the map

N -> {0, ..., p-1} mapping i to 2^i mod p.

It immediately follow from Euler's theorem that 2^i mod p = 2^j mod p if i is congruent to j modulo phi(p), where phi denotes Euler's totient function.

Therefore, the map is periodic with a period at most phi(p) = p-1. But for which primes is equality achieved? It happens quite often, for instance for p = 3, 5, 11, 13, 19 the period of the map is indeed p-1, but not for p=7 where the map has only period three: 2, 4, 1, 2, 4, 1, ....

(It seems p=7 being Mersenne is a hint?) What is the general criterion on p?

Second: assume p is such a prime such that the period is the maximum p-1 allowed by Euler's theorem. We then get a permutation on the set of p-1 elements, defined by restricting our map to {1, ... , p-1} . Indeed, this map does not hit zero because 2^i mod p = 0 would mean that 2^i = k * p for some k > 0 but p was supposed to be a prime > 2 so it can't occur in the decomposition of 2^i, and the injectivity follows from the assumed "maximal period" property. Also, Fermat's little theorem says that the permutation takes the value 1 at p-1.

Is there anything else interesting to say about this permutation? It starts off with all powers of two <= p-1, and then fills all gaps between them. Does it do so in a way that has strong properties (any symmetries, for instance) or not? Sorry for the this part of the question being a little vague. Thanks for nice answers.

Edit: apparently, these permutations are closely related to so-called Costas permutations which have no symmetries at all in some sense. Nevertheless, the second part of the question is still open.

Heard they've been doing some shady business.

BamBOO!

They’re on standbi

Pilot on me!!

Nothing, he was gladiator.

Christopher Walken

Dad jokes are supposed to be jokes you can tell a kid and they will understand it and find it funny.

This sub is mostly just NSFW puns now.

If it needs a NSFW tag it's not a dad joke. There should just be a NSFW puns subreddit for that.

Edit* I'm not replying any longer and turning off notifications but to all those that say "no one cares", there sure are a lot of you arguing about it. Maybe I'm wrong but you people don't need to be rude about it. If you really don't care, don't comment.

What did 0 say to 8 ?

" Nice Belt "

So What did 3 say to 8 ?

" Hey, you two stop making out "

When I got home, they were still there.

I won't be doing that today!