

My favorite math problem

Introduction According to legend, the imaginary unit i = √-1 was invented by mathematicians because the equation x2 + 1 = 0 had no solution. So they introduced a new number i, defined so that i2 + 1 = 0, and investigated the consequences. Like most legends, this one has very little to do with the real history of the thing it purports to describe. But let's pretend that it's true for a while. Descartes' theorem If P(x) is some polynomial, and z is some number such that P(z) = 0, then we say that z is a zero of P or a root of P. Descartes' theorem (yes, that Descartes) says that if z is a zero of P, then P can be factored into P = (x-z)P', where P' is a polynomial of lower degree, and has all the same roots as P, except without z. For example, consider P = x3 - 6x2 + 11x - 6. One of the zeroes of P is 1, so Descartes theorem says that we can write P in the form (x-1)P' for some second-degree polynomial P', and if there are any other zeroes of P, they will also be zeroes of P'. And in fact we can; x3 - 6x2 + 11x - 6 = (x-1)(x2 - 5x + 6). The other zeroes of the original polynomial P are 2 and 3, and both of these (but not 1) are also zeroes of x2 - 5x + 6. We can repeat this process. If a 1 , a 2 , ... a n are the zeroes of some nth-degree polynomial P, we can factor P as (x-a 1 )(x-a 2 )...(x-a n ). The only possibly tricky thing here is that some of the a i might be the same. x2 - 2x + 1 has a zero of 1, but twice, so it factors as (x-1)(x-1). x3 - 4x2 + 5x - 2 has a zero of 1, but twice, so it factors as (x-1)(x2 - 3x + 2), where (x2 - 3x + 2) also has a zero of 1, but this time only once. Descartes' theorem has a number of important corollaries: an nth-degree polynomial has no more than n zeroes. A polynomial that has a zero can be factored. When coupled with the so-called fundamental theorem of algebra, which says that every polynomial has a zero over the complex numbers, it implies that every nth-degree polynomial can be factored into a product of n factors of the form (x-z) where z is complex, or into a product of first- and second-degree polynomials with real coefficients. The square roots of -1 Suppose we want to know what the square roots of 1 are. We want numbers x such that x2 = 1, or, equivalently, such that x2 - 1 = 0. x2 - 1 factors into (x-1)(x+1), so the zeroes are ±1, and these are the square roots of 1. On the other hand, if we want the square roots of 0, we need the solutions of x2 = 0, and the corresponding calculation calls for us to factor x2, which is just (x-0)(x-0), So 0 has the same square root twice; it is the only number without two distinct square roots. If we want to find the square roots of -1, we need to factor x2 + 1. There are no real numbers that do this, so we call the polynomial irreducible. Instead, we make up some square roots for it, which we will call i and j. We want (x - i)(x - j) = x2 + 1. Multiplying through gives x2 - (i + j)x + ij = x2 + 1. So we have some additional relationships between i and j: ij = 1, and i + j = 0. Because i + j = 0, we always dispense with j and just call it -i. But it's essential to realize that neither one is more important than the other. The two square roots are identical twins, different individuals that are completely indistinguishable from each other. Our choice about which one was primary was completely arbitrary. That doesn't even do justice to the arbitrariness of the choice: i and j are so indistinguishable that we still don't know which one got the primary notation. Every property enjoyed by i is also enjoyed by j. i3 = j, but j3 = i. eiπ = -1, but so too ejπ = -1. And so on. In fact, I once saw the branch of mathematics known as Galois theory described as being essentially the observation that i and -i were indistinguishable, plus the observation that there are no other indistinguishable pairs of complex numbers. Anyway, all this is to set up the main point of the article, which is to describe my favorite math problem of all time. I call it the "train problem" because I work on it every time I'm stuck on a long train trip. We are going to investigate irreducible polynomials; that is, polynomials that do not factor into products of simpler polynomials. Descartes' theorem implies that irreducible polynomials are zeroless, so we are going to start by looking for polynomials without zeroes. But we are not going to do it in the usual world of the real numbers. The answer in that world is simple: x2 + 1 is irreducible, and once you make up a zero for this polynomial and insert it into the universe, you are done; the fundamental theorem of algebra guarantees that there are no more irreducible polynomials. So instead of the reals, we're going to do this in a system known as Z 2 . Z 2 Z 2 is much simpler than the real numbers. Instead of lots and lots of numbers, Z 2 has only two. (Hence the name: The 2 means 2 and the Z means "numbers". Z is short for Zahlen, which is German.) Z 2 has only 0 and 1. Addition and multiplication are as you expect, except that instead of 1 + 1 = 2, we have 1 + 1 = 0. + 0 1 0 0 1 1 1 0 × 0 1 0 0 0 1 0 1 The 1+1=0 oddity is the one and only difference between Z 2 and ordinary algebra. All the other laws remain the same. To be explicit, I will list them below: a + 0 = a

+ 0 = a + b = b + a

+ = + ( a + b ) + c = a + ( b + c )

+ ) + = + ( + ) a × 0 = 0

× 0 = 0 a × 1 = a

× 1 = a × b = b × a

× = × ( a × b ) × c = a × ( b × c )

× ) × = × ( × ) a × (b + c) = a × c + b × c Algebra in Z 2 But the 1+1=0 oddity does have some consequences. For example, a + a = a·(1 + 1) = a·0 = 0 for all a. This will continue to be true for all a, even if we extend Z 2 to include more elements than just 0 and 1, just as x + x = 2x is true not only in the real numbers but also in their extension, the complex numbers. Because a + a = 0, we have a = -a for all a, and this means that in Z 2 , addition and subtraction are the same thing. If you don't like subtraction—and really, who does?—then Z 2 is the place for you. As a consequence of this, there are some convenient algebraic manipulations. For example, suppose we have a + b = c and we want to solve for a. Normally, we would subtract b from both sides. But in Z 2 , we can add b to both sides, getting a + b + b = c + b. Then the b + b on the left cancels (because b + b = 0) leaving a = c + b. From now on, I'll call this the addition law. If you suffered in high school algebra because you wanted (a + b)2 to be equal to a2 + b2, then suffer no more, because in Z 2 they are equal. Here's the proof: (a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b) = (a2 + ab) + (ba + b2) = a2 + ab + ab + b2 = a2 + b2 Because the ab + ab cancels in the last step. This happy fact, which I'll call the square law, also holds true for sums of more than two terms: (a + b + ... + z)2 = a2 + b2 + ... + z2. But it does not hold true for cubes; (a + b)3 is a3 + a2b + ab2 + b3, not a3 + b3. Irreducible polynomials in Z 2 Now let's try to find a polynomial that is irreducible in Z 2 . The first degree polynomials are x and x+1, and both of these have zeroes: 0 and 1, respectively. There are four second-degree polynomials: x2 x2 + 1 x2 + x x2 + x + 1 If a second-degree polynomial is reducible, it must factor into a product of two first-degree polynomials, and there are three such products: x·x, (x+1)(x+1), and x(x+1). These are, respectively, polynomials 1, 2, and 3 in the previous display. The polynomials have zeroes as indicated here: Polynomial Zeroes Factorization x2 0 (twice) x2 x2 + 1 1 (twice) (x+1)2 x2 + x 0, 1 x(x+1) x2 + x + 1 none Note in particular that, unlike in the real numbers, x2 + 1 is not irreducible in Z 2 . 1 is a zero of this polynomial, which, because of the square law, factors as (x+1)(x+1). But we have found an irreducible polynomial, x2 + x + 1. The other three second-degree polynomials all can be factored, but x2 + x + 1 cannot. Like the fictitious mathematicians who invented i, we will rectify this problem, by introducing into Z 2 a new number, called b, which has the property that b2 + b + 1 = 0. What are the consequences of this? The first thing to notice is that we haven't added only one new number to Z 2 . When we add i to the real numbers, we also get a lot of other stuff like i+1 and 3i-17 coming along for the ride. Here we can't add just b. We must also add b+1. The next thing we need to do is figure out how to add and multiply with b. The addition and multiplication tables are extended as follows: + 0 1 b b+1 0 0 1 b b+1 1 1 0 b+1 b b b b+1 0 1 b+1 b+1 b 1 0 × 0 1 b b+1 0 0 0 0 0 1 0 1 b b+1 b 0 b b+1 1 b+1 0 b+1 1 b The only items here that might require explanation are the entries in the lower-right-hand corner of the multiplication table. Why is b×b = b+1? Well, b2 + b + 1 = 0, by definition, so b2 = b + 1 by the addition law. Why is b × (b+1) = 1? Well, b × (b+1) = b2 + b = (b + 1) + b = 1 because the two b's cancel. Perhaps you can work out (b+1) × (b+1) for yourself. From here there are a number of different things we could investigate. But what I think is most crucial is to discover the value of b3. Let's calculate that. b3 = b·b2. b2, we know, is b+1. So b3 = b(b+1) = 1. Thus b is a cube root of 1. People with some experience with this sort of study will not be surprised by this. The cube roots of 1 are precisely the zeroes of the polynomial x3 + 1 = 0. 1 is obviously a cube root of 1, and so Descartes' theorem says that x3 + 1 must factor into (x+1)P, where P is some second-degree polynomial whose zeroes are the other two cube roots. P can't be divisible by x, because 0 is not a cube root of 1, so it must be either (x+1)(x+1), or something else. If it were the former, then we would have x3 + 1 = (x+1)3, but as I pointed out before, we don't. So it must be something else. The only candidate is x2 + x + 1. Thus the other two cube roots of 1 must be zeroes of x2 + x + 1. The next question we must answer without delay is: b is one of the zeroes of x2 + x + 1; what is the other? There is really only one possibility: it must be b+1, because that is the only other number we have. But another way to see this is to observe that if p is a cube root of 1, then p2 must also be a cube root of 1: b3 = 1 (b3)2 = 12 b6 = 1 (b2)3 = 1 So we now know the three zeroes of x3 + 1: they are 1, b, and b2. Since x3 + 1 factors into (x+1)(x2+x+1), the two zeroes of x2+x+1 are b and b2. As we noted before, b2 = b+1. This means that x2 + x + 1 should factor into (x+b)(x+b+1). Multiplying through we get x2 + bx + (b+1)x + b(b+1) = x2 + x + 1, as hoped. Actually, the fact that both b and b2 are zeroes of x2+x+1 is not a coincidence. Z 2 has the delightful property that if z is a zero of any polynomial whose coefficients are all 0's and 1's, then z2 is also a zero of that polynomial. I'll call this Theorem 1. Theorem 1 is not hard to prove; it follows from the square law. Feel free to skip the demonstration which follows in the rest of the paragraph. Suppose the polynomial is P(x) = a n xn + a n-1 xn-1 + ... + a 1 x + a 0 , and z is a zero of P, so that P(z) = 0. Then (P(z))2 = 0 also, so (a n zn + a n-1 zn-1 + ... + a 1 z + a 0 )2 = 0. By the square law, we have a n 2(zn)2 + a n-1 2(zn-1)2 + ... + a 1 2z2 + a 0 2 = 0. Since all the a i are either 0 or 1, we have a i 2 = a i , so a n (zn)2 + a n-1 (zn-1)2 + ... + a 1 z2 + a 0 = 0. And finally, (zk)2 = (z2)k, so we get a n (z2)n + a n-1 (z2)n-1 + ... + a 1 z2 + a 0 = P(z2) = 0, which is what we wanted to prove. Now you might be worried about something: the fact that b is a zero of x2 + x + 1 implies that b2 is also a zero; doesn't the fact that b2 is a zero imply that b4 is also a zero? And then also b8 and b16 and so forth? Then we would be in trouble, because x2 + x + 1 is a second-degree polynomial, and, by Descartes' theorem, cannot have more than two zeroes. But no, there is no problem, because b3 = 1, so b4 = b·b3 = b·1 = b, and similarly b8 = b2, so we are getting the same two zeroes over and over. Symmetry again Just as i and -i are completely indistinguishable from the point of view of the real numbers, so too are b and b+1 indistinguishable from the point of view of Z 2 . One of them has been given a simpler notation than the other, but there is no way to tell which one we gave the simpler notation to! The one and only defining property of b is that b2 + b + 1 = 0; since b+1 also shares this property, it also shares every other property of b. b and b+1 are both cube roots of 1. Each, added to the other gives 1; each, added to 1 gives the other. Each, when multiplied by itself gives the other; each when multiplied by the other gives 1. Third-degree polynomials We have pretty much exhausted the interesting properties of b and its identical twin b2, so let's move on to irreducible third-degree polynomials. If a third-degree polynomial is reducible, then it is either a product of three linear factors, or of a linear factor and an irreducible second-degree polynomial. If the former, it must be one of the four products x3, x2(x+1), x(x+1)2, or (x+1)3. If the latter, it must be one of x(x2+x+1) or (x+1)(x2+x+1). This makes 6 reducible polynomials. Since there are 8 third-degree polynomials in all, the other two must be irreducible: Polynomial Zeroes Factorization x3 0, 0, 0 x3 x3 + 1 1, b, b+1 (x+1)(x2+x+1) x3 + x 0, 1, 1 x(x+1)2 x3 + x + 1 none x3 + x2 0, 0, 1 x2(x+1) x3 + x2 + 1 none x3 + x2 + x 0, b, b+1 x(x2+x+1) x3 + x2 + x + 1 1, 1, 1 (x+1)3 There are two irreducible third-degree polynomials. Let's focus on x3 + x + 1, since it seems a little simpler. As before, we will introduce a new number, c, which has the property c3 + c + 1 = 0. By the addition law, this implies that c3 = c + 1. Using this, we can make a table of powers of c, which will be useful later: c1 = c c2 = c2 c3 = c + 1 c4 = c2 + c c5 = c2 + c + 1 c6 = c2 + 1 c7 = 1 To calculate the entries in this table, just multiply each line by c to get the line below. For example, once we know that c5 = c2 + c + 1, we multiply by c to get c6 = c3 + c2 + c. But c3 = c + 1 by definition, so c6 = (c + 1) + c2 + c = c2 + 1. Once we get as far as c7, we find that c is a 7th root of 1. Analogous to the way b and b2 were both cube roots of 1, the other 7th roots of 1 are c2, c3, c4, c5, c6, and 1. For example, c5 is a 7th root of 1 because (c5)7 = c35 = (c7)5 = 15 = 1. We've decided that c is a zero of x3 + x + 1. What are the other two zeroes? By theorem 1, they must be c2 and (c2)2 = c4. Wait, what about c8? Isn't that a zero also, by the same theorem? Oh, but c8 = c·c7 = c, so it's really the same zero again. What about c3, c5, and c6? Those are the zeroes of the other irreducible third-degree polynomial, x3 + x2 + 1. For example, (c3)3 + (c3)2 + 1 = c9 + c6 + 1 = c2 + (c2+1) + 1 = 0. So when we take Z 2 and adjoin a new number that is a solution of the previously unsolvable equation x3 + x + 1 = 0, we're actually forced to add six new numbers; the resulting system has 8, and includes three different solutions to both irreducible third-degree polynomials, and a full complement of 7th roots of 1. Since the zeroes of x3 + x + 1 and x3 + x2 + 1 are all 7th roots of 1, this tells us that x7 + 1 factors as (x+1)(x3 + x + 1)(x3 + x2 + 1), which might not have been obvious. Onward I said this was my favorite math problem, but I didn't say what the question was. But by this time we have dozens of questions. For example: We've seen cube roots and seventh roots of 1. Where did the fifth roots go? (The square theorem ensures that even-order roots are uninteresting. The fourth roots of 1, for example, are just 1, 1, 1, and 1. The sixth roots are the same as the cube roots, but repeated: 1, 1, b , b , b 2 , and b 2 .) The fifth roots, it turns out, appear later, as the four zeroes of the polynomial x 4 + x 3 + x 2 + x + 1. These four, and 1, are five of the 15 fifteenth roots of 1. Eight others appear as the zeroes of the other two irreducible fourth-degree polynomials, and the other two fifteenth roots are b and b 2 . If we let d be a zero of one of the other irreducible fourth-degree polynomials, say x 4 + x + 1, then the zeroes of this polynomial are d , d 2 , d 4 , and d 8 , and the zeroes of the other are d 7 , d 11 , d 13 , and d 14 . So we have an interesting situation. We can extend Z 2 by adjoining the zeroes of x 3 + 1, to get a larger system with 4 elements instead of only 2. And we can extend Z 2 by adjoining the zeroes of x 7 + 1, to get a larger system with 8 elements instead of only 2. But we can't do the corresponding thing with x 5 + 1, because the fifth roots of 1 don't form a closed system. The sum of any two of the seventh roots of 1 was another seventh root of 1; for example, c 3 + c 4 = c 6 and c + c 2 = c 4 . But the corresponding fact about fifth roots is false. Suppose, for example, that y 5 = 1. We'd like for y + y 2 to be another fifth root of 1, but it isn't; it actually turns out that ( y + y 2 ) 5 = b . For which n do the n th roots of 1 form a closed system?

What do we get if we extend Z 2 with both b and c ? What's bc ? What's b + c ? (Spoiler: b + c turns out to be zero of x 6 + x 5 + x 3 + x 2 + 1, and is a 63rd root of 1, so we may as well call it f . bc = f 48 .) The degree of a number is the degree of the simplest polynomial of which it is a zero. What's the degree of b + c ? In general, if p and q have degrees d ( p ) and d ( q ), respectively, what can we say about the degrees of pq and p + q ?

with both and ? What's ? What's + ? (Spoiler: + turns out to be zero of + + + + 1, and is a 63rd root of 1, so we may as well call it . = .) Since there is an n th root of 1 for every odd n , that implies that for all odd n , there is some k such that n divides 2 k - 1. How are n and k related?

th root of 1 for every odd , that implies that for all odd , there is some such that divides 2 - 1. How are and related? For n = 2, 3, and 4, the polynomial x n + x + 1 was irreducible. Is this true in general? (No: x 5 + x + 1 = ( x 3 + x 2 + 1)( x 2 + x + 1).) When is it true? Is there a good way to find irreducible polynomials? How many irreducible polynomials are there of degree n ? (The sequence starts out 0, 0, 1, 2, 3, 6, 9, 18, 30, 56, 99,...)

= 2, 3, and 4, the polynomial + + 1 was irreducible. Is this true in general? (No: + + 1 = ( + + 1)( + + 1).) When is it true? Is there a good way to find irreducible polynomials? How many irreducible polynomials are there of degree ? (The sequence starts out 0, 0, 1, 2, 3, 6, 9, 18, 30, 56, 99,...) The Galois theorem says that any finite field that contains Z 2 must have 2 n elements, and its addition component must be isomorphic to a direct product of n copies of Z 2 . Extending Z 2 with c , for example, gives us the field GF (8), obtained as the direct product of {0, 1}, {0, c }, and {0, c 2 }. The addition is very simple, but the multiplication, viewed in this light, is rather difficult, since it depends on knowing the secret identity c · c 2 = c + 1. Identify the element a 0 + a 1 c + a 2 c 2 with the number a 0 + 2 a 1 + 4 a 2 , and let the addition operation be exclusive-or. What happens to the multiplication operation?

must have 2 elements, and its addition component must be isomorphic to a direct product of copies of . Extending with , for example, gives us the field (8), obtained as the direct product of {0, 1}, {0, }, and {0, }. The addition is very simple, but the multiplication, viewed in this light, is rather difficult, since it depends on knowing the secret identity · = + 1. Consider a linear-feedback shift register: # Arguments: N, a positive integer # 0 <= K <= 2**N-1 MAXBIT = 2**N c = 2 until c == 1: c *= 2 if (c => MAXBIT): c &= ~MAXBIT c ^= K print c (Except that this function runs backward relative to the way that LFSRs usually work.) Does this terminate for all choices of N and K ? (Yes.) For any given N , the maximum number of iterations of the until loop is 2 N - 1. For which choices of K is this maximum achieved?

(Except that this function runs backward relative to the way that LFSRs usually work.) Consider irreducible sixth-degree polynomials. We should expect that adjoining a zero of one of these, say f , will expand Z 2 to contain 64 elements: 0, 1, f , f 2 , ..., f 62 , and f 63 = 1. In fact, this is the case. But of these 62 new elements, some are familiar. Specifically, f 9 = c , and f 21 = b . Of the 62 powers of f , 8 are numbers we have seen before, leaving only 54 new numbers. Each irreducible 6th-degree polynomial has 6 of these 54 as its zeroes, so there are 9 such polynomials. What does this calculation look like in general?

, will expand to contain 64 elements: 0, 1, , , ..., , and = 1. In fact, this is the case. c 1 , c 2 , and c 4 are zeroes of x 3 + x + 1, one of the two irreducible third-degree polynomials. This means that x 3 + x + 1 = ( x + c 1 )( x + c 2 )( x + c 4 ). Multiplying out the right-hand side, we get: x 3 + ( c + c 2 + c 4 ) x 2 + ( c 3 + c 5 + c 6 ) x + 1. Equating coefficients, we have c + c 2 + c 4 = 0 and c 3 + c 5 + c 6 = 1. c 3 , c 5 , and c 6 are the zeroes of the other irreducible third-degree polynomial. Coincidence? (No.)

, , and are zeroes of + + 1, one of the two irreducible third-degree polynomials. This means that + + 1 = ( + )( + )( + ). Multiplying out the right-hand side, we get: + ( + + ) + ( + + ) + 1. Equating coefficients, we have + + = 0 and + + = 1. , , and are the zeroes of the irreducible third-degree polynomial. Coincidence? (No.) The irreducible polynomials appear to come in symmetric pairs: if a n x n + a n -1 x n -1 + ... + a 1 x + a 0 is irreducible, then so is a 0 x n + a 1 x n -1 + ... + a n -1 x + a n . Why?

+ + ... + + is irreducible, then so is + + ... + + . Why? c 1 , c 2 , and c 4 are the zeroes of one of the irreducible third-degree polynomials. 1, 2, and 4 are all the three-bit binary numbers that have exactly one 1 bit. c 3 , c 5 , and c 6 are the zeroes the other irreducible third-degree polynomial. 3, 5, and 6 are all the three-bit binary numbers that have exactly one 0 bit. Coincidence? (No; theorem 1 is closely related here.) Now consider the analogous case for fourth-degree polynomials. The 4-bit numbers that have one 1 bit are 0001, 0010, 0100, and 1000 (1, 2, 4, and 8) which are all zeroes of one of the irreducible 4th-degree polynomials. The 4-bit numbers that have one 0 bit are 1110, 1101, 1011, and 0111 (7, 11, 13, and 14) which are all zeroes of another of the irreducible 4th-degree polynomials. The 4-bit numbers with two 1 bits and two 0 bits fall into two groups. In one group we have the numbers where similar bits are adjacent: 0011, 0110, 1100, and 1001, or 3, 6, 9, and 12. These are the zeroes of the third irreducible 4th-degree polynomial. The remaining 4-bit numbers are 0101 and 1010, which correspond to b and b 2 . There is, therefore, a close relationship between the structure an extension of Z 2 and the so-called "necklace patterns" of bits, which are patterns that imagine that the bits are joined into a circle with no distinguishable starting point. The exceptional values ( b and b 2 ) corresponded to necklace patterns that did not possess the full n -fold symmetry. This occurs for composite values of n . But we might also expect to find exceptions at n = 11, since then 2 n -1 = 2047 is composite. However, 2047 is exceptional in another way: 2047 = 23 · 89, and neither 23 nor 89 is a divisor of 2 n -1 for any smaller value of n . Can we prove from this that when p is prime and 2 p -1 is not, the divisors of 2 p -1 do not divide 2 q -1 for any q < p ?

, , and are the zeroes of one of the irreducible third-degree polynomials. 1, 2, and 4 are all the three-bit binary numbers that have exactly one 1 bit. , , and are the zeroes the other irreducible third-degree polynomial. 3, 5, and 6 are all the three-bit binary numbers that have exactly one 0 bit. Coincidence? (No; theorem 1 is closely related here.) There is, therefore, a close relationship between the structure an extension of and the so-called "necklace patterns" of bits, which are patterns that imagine that the bits are joined into a circle with no distinguishable starting point. Since ci are the seventh roots of 1, we have x7 + 1 = (x + 1)(x + c)(x + c2)...(x + c6). After dividing out the trivial factor of x+1, we are left with: x6 + x5 + ... + 1 = (x + c)(x + c2)...(x + c6) Multiplying through on the right side and equating coefficients gives: c + c2 + ... + c6 = 1 cc2 + cc3 + ... + c5c6 = 1 cc2c3 + cc2c4 + ... + c4c5c6 = 1 ... = ... cc2c3c4c5c6 = 1 What, if anything, can one make of this? The reason this problem works so well as a long-train-trip problem is that it can be approached from many different directions and at many different levels of difficulty. If I am tired and woozy, I can still enumerate fifth-degree polynomials looking for the irreducible ones, calculate tables of values of en, and ponder the interrelationships of the exponents until I fall asleep. When I wake up again, rested, I can consider the deep relationships with the Galois theorem, with algebra, with random number generation, and with cryptography. And then, when I get tired of Z 2 , I can start all over with Z 3 .

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