EXTRA CREDIT FOR MATH 39

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1 EXTRA CREDIT FOR MATH 39 This is the second, theoretical, part of an extra credit homework. This homework in not compulsory. If you do it, you can get up to 6 points (3 points for each part) of extra credit on your average grade for the class, over 100. This presents the solution of the first part of the extra credit, and ask you to solve some questions to get the three more bonus points. The questions to be solved are in and Empirical study The first part is an empirical study of some number-theoretic properties. You will need to use a computer, and more precisely either a programming language (e.g, C, C++, Java, python, anything you like) or a mathematical software (mathematica, maple, magma, sage, etc.). Your work will be a little bit easier if you know or learn how to use a mathematical software. For the first part, you are authorized to work by pairs and to handle only one set of answers for every pair. The two members of a pair will get the same grade. If you need help, please tell me. You are required to join a printed listing of your program with your work First digit of big numbers. 1. Look up (on the internet or wherever you want) the population of the 50 states of the U.S. (or 51 including D.C). Count how many times those numbers begin by the digit 1, by the digit 2, and so on until the digit 9. Do the same for other variables: the area of the states (in square miles, or in square kilometers as you wish), their GDP. Then, for each of the series of digits you got, compute the proportion of 1 s, of 2 s, etc. In each series, which digit is the most frequent? Did you expect that? 2. With the help of a computer, compute the first digit of 2 n (in decimal writing) for n = 1 to n as large as you can. Compute the proportion of 1 s, of 2 s, etc in this sequence of digits. Do the same with 2 n replaced by 3 n, 5 n, 6 n. Can you make a conjecture? 1.2. Higher reciprocity law. 1. Let P (X) = X 3 2, Q(X) = X 7 1 and R(X) = X 3 + X 2 2X 1. For each of the 200 first prime numbers p, compute the number of roots of the polynomials P (X), Q(X), R(X) modulo p. For each of those polynomials, can you make conjectures on a. the proportion of primes for which the polynomial has a given number of roots modulo p? b. a simple rule on p that will predict how many roots the polynomial has modulo p? (but be aware that this might not be possible for all three polynomials) 1

2 2 EXTRA CREDIT FOR MATH Solutions and theoretical questions 2.1. First digit of big numbers Statements. I won t provide a solution to exercise 1, see: s_law For exercise 2, experiments with the sequence 2 n, 3 n, etc. show that many more of those numbers begin by a 1 (about 30%) than by a 9 (only about 5%), and in general, that the proportion of numbers in the sequence that begins by a given digit is decreasing with the digit. It might have been difficult, without knowing Benford s law, to guess what the proportion of numbers (in the sequence 2 n, or 3 n, or a n for any a which is not a power of 10) that begins by 1, 2,,... converge to, but the answer is P (1) = log 10 (2) log 10 (1) = log 10 (2), P (2) = log 10 (3) log 10 (2), and in general, P (d) = log 10 (d + 1) log 10 (d). The precise mathematical statement is: Theorem 1. Let a be a natural number that is not a power of 10. Let d be a digit (a number between 1 and 9). For N any integer, let b d (N) the number of integers a n with 0 n N that begins by the digit d. Then lim b d(n)/n = P (d). N Questions. 1. Begin by checking that P (1) + P (2) + + P (9) = 1, which is what we expect for proportions that have to add up to Show that the first digit of an integer M is M/10 log 10 M. 3. Show that the first digit d of a n is 10 n log 10 a n log 10 a. In particular, the digit d is determined by the number r = n log 10 a n log 10 a as follows: d = 1 if 0 r < log 10 2 d = 2 if log 10 2 r < log 10 3,... d = 9 if log 10 9 r < log = Show that log 10 a is an irrational number (under the assumption on a of Theorem 1). The equipartition theorem theorem states: Let 0 c d 1. Let x be an irrational number. For every N, let p(n) be the number of integers n, 0 n N, such that nx nx is between c an d (it doesn t matter is between is understood strictly, that is including c and d, or not). Then lim p(n)/n = d c. N 5. Using the equipartition theorem, prove Theorem Higher reciprocity law.

3 EXTRA CREDIT FOR MATH Solutions. Let us consider first the case of P (X) = X 3 2. If p is a prime, the number of possible incongruent roots for P (X) (mod p) is at most 3, as seen in class. Here is a program in Sage 1 that compute and print that for the prime up to 100. This is enough to make conjectures on question b, and more would be a waste of paper. In order to help solving question b, and to see if there is a rule that predicts in term of a congruences of p modulo some modulus the number of roots of P (X) mod p, the residue of p modulo various small modulus are also written. The program also compute the proportion of primes for which the polynomial P (X) has 0, 1, 2 or 3 roots. To make the answer appear more clearly, the programs compute this proportion for primes up to 100, then 2100, etc. print number of roots of X^3-2 mod p for n in range(10): max= *n l=list(primes(max)) result=[0,0,0,0]; for p in l: j=0; for x in range(p): if p.divides(x^3-2): j=j+1; result[j] = result[j]+1; if n==0: print p=,p, number of roots=,j, p mod 3,4,5,6,7=, p%3, p%4,p%5,p% r=sum(result); print print( For the primes up to,max) for i in range(4): print proportion of primes with,i, roots:,result[i]/r+0.0; number of roots of X^3-2 mod p p= 2 number of roots= 1 p mod 3,4,5,6,7= p= 3 number of roots= 1 p mod 3,4,5,6,7= p= 5 number of roots= 1 p mod 3,4,5,6,7= p= 7 number of roots= 0 p mod 3,4,5,6,7= p= 11 number of roots= 1 p mod 3,4,5,6,7= p= 13 number of roots= 0 p mod 3,4,5,6,7= p= 17 number of roots= 1 p mod 3,4,5,6,7= p= 19 number of roots= 0 p mod 3,4,5,6,7= p= 23 number of roots= 1 p mod 3,4,5,6,7= p= 29 number of roots= 1 p mod 3,4,5,6,7= Sage is free software of mathematical computations. You can use it online at The syntax is based on the language python

4 4 EXTRA CREDIT FOR MATH 39 p= 31 number of roots= 3 p mod 3,4,5,6,7= p= 37 number of roots= 0 p mod 3,4,5,6,7= p= 41 number of roots= 1 p mod 3,4,5,6,7= p= 43 number of roots= 3 p mod 3,4,5,6,7= p= 47 number of roots= 1 p mod 3,4,5,6,7= p= 53 number of roots= 1 p mod 3,4,5,6,7= p= 59 number of roots= 1 p mod 3,4,5,6,7= p= 61 number of roots= 0 p mod 3,4,5,6,7= p= 67 number of roots= 0 p mod 3,4,5,6,7= p= 71 number of roots= 1 p mod 3,4,5,6,7= p= 73 number of roots= 0 p mod 3,4,5,6,7= p= 79 number of roots= 0 p mod 3,4,5,6,7= p= 83 number of roots= 1 p mod 3,4,5,6,7= p= 89 number of roots= 1 p mod 3,4,5,6,7= p= 97 number of roots= 0 p mod 3,4,5,6,7= ( For the primes up to, 100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 2100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 4100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 6100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 8100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots:

5 EXTRA CREDIT FOR MATH 39 5 ( For the primes up to, 10100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 12100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 14100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 16100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 18100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: We know turn to the polynomial Q(X) = X 7 1. Here the program has to be slightly modified since X 7 1 may have up to 7 roots modulo p print number of roots of X^7-1 mod p for n in range(10): max= *n l=list(primes(max)) result=[0,0,0,0,0,0,0,0]; for p in l: j=0; for x in range(p): if p.divides(x^7-1): j=j+1; result[j] = result[j]+1; if n==0: print p=,p, number of roots=,j, p mod 7=, p%7; r=sum(result);

6 6 EXTRA CREDIT FOR MATH 39 print print( For the primes up to,max) for i in range(8): print proportion of primes with,i, roots:,result[i]/r+0.0; number of roots of X^7-1 mod p p= 2 number of roots= 1 p mod 7= 2 p= 3 number of roots= 1 p mod 7= 3 p= 5 number of roots= 1 p mod 7= 5 p= 7 number of roots= 1 p mod 7= 0 p= 11 number of roots= 1 p mod 7= 4 p= 13 number of roots= 1 p mod 7= 6 p= 17 number of roots= 1 p mod 7= 3 p= 19 number of roots= 1 p mod 7= 5 p= 23 number of roots= 1 p mod 7= 2 p= 29 number of roots= 7 p mod 7= 1 p= 31 number of roots= 1 p mod 7= 3 p= 37 number of roots= 1 p mod 7= 2 p= 41 number of roots= 1 p mod 7= 6 p= 43 number of roots= 7 p mod 7= 1 p= 47 number of roots= 1 p mod 7= 5 p= 53 number of roots= 1 p mod 7= 4 p= 59 number of roots= 1 p mod 7= 3 p= 61 number of roots= 1 p mod 7= 5 p= 67 number of roots= 1 p mod 7= 4 p= 71 number of roots= 7 p mod 7= 1 p= 73 number of roots= 1 p mod 7= 3 p= 79 number of roots= 1 p mod 7= 2 p= 83 number of roots= 1 p mod 7= 6 p= 89 number of roots= 1 p mod 7= 5 p= 97 number of roots= 1 p mod 7= 6 ( For the primes up to, 100) proportion of primes with 1 roots: proportion of primes with 7 roots: ( For the primes up to, 2100)

7 EXTRA CREDIT FOR MATH 39 7 proportion of primes with 1 roots: proportion of primes with 7 roots: ( For the primes up to, 4100) proportion of primes with 1 roots: proportion of primes with 7 roots: ( For the primes up to, 6100) proportion of primes with 1 roots: proportion of primes with 7 roots: ( For the primes up to, 8100) proportion of primes with 1 roots: proportion of primes with 7 roots: ( For the primes up to, 10100) proportion of primes with 1 roots:

8 8 EXTRA CREDIT FOR MATH 39 proportion of primes with 7 roots: ( For the primes up to, 12100) proportion of primes with 1 roots: proportion of primes with 7 roots: ( For the primes up to, 14100) proportion of primes with 1 roots: proportion of primes with 7 roots: ( For the primes up to, 16100) proportion of primes with 1 roots: proportion of primes with 7 roots: ( For the primes up to, 18100) proportion of primes with 1 roots: proportion of primes with 7 roots: Finally, here are the result for the polynomial R(X) (we omit the program which is similar to the above ones) number of roots of X^3+X^2-2X-1 mod p

9 p= 2 number of roots= 0 p mod 7= 2 p= 3 number of roots= 0 p mod 7= 3 p= 5 number of roots= 0 p mod 7= 5 p= 7 number of roots= 1 p mod 7= 0 p= 11 number of roots= 0 p mod 7= 4 p= 13 number of roots= 3 p mod 7= 6 p= 17 number of roots= 0 p mod 7= 3 p= 19 number of roots= 0 p mod 7= 5 p= 23 number of roots= 0 p mod 7= 2 p= 29 number of roots= 3 p mod 7= 1 p= 31 number of roots= 0 p mod 7= 3 p= 37 number of roots= 0 p mod 7= 2 p= 41 number of roots= 3 p mod 7= 6 p= 43 number of roots= 3 p mod 7= 1 p= 47 number of roots= 0 p mod 7= 5 p= 53 number of roots= 0 p mod 7= 4 p= 59 number of roots= 0 p mod 7= 3 p= 61 number of roots= 0 p mod 7= 5 p= 67 number of roots= 0 p mod 7= 4 p= 71 number of roots= 3 p mod 7= 1 p= 73 number of roots= 0 p mod 7= 3 p= 79 number of roots= 0 p mod 7= 2 p= 83 number of roots= 3 p mod 7= 6 p= 89 number of roots= 0 p mod 7= 5 p= 97 number of roots= 3 p mod 7= 6 EXTRA CREDIT FOR MATH 39 9 ( For the primes up to, 100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 2100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 4100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 6100) proportion of primes with 0 roots:

10 10 EXTRA CREDIT FOR MATH 39 proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 8100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 10100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 12100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 14100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 16100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: ( For the primes up to, 18100) proportion of primes with 0 roots: proportion of primes with 1 roots: proportion of primes with 3 roots: Observations. Here we study the data given above, looking for some patterns, and we record what we find, as unproved statements (that may be false until proof that they are correct) We begin by some easy observations: (i) The number of roots of P (X) (mod p) can be 0, 1 or 3 but never 2. (ii) The number of roots of Q(X) (mod p) can only be 1 and 7, but never 0, or between 2 and 6.

11 EXTRA CREDIT FOR MATH (iii) The number of roots of R(X) (mod p) can only be 0 or 3 excepted for p = 7 for which it is 1. The next one needs to be a little bit more attentive (iv) If the number of roots of Q(X) (mod p) is 7, then the number of roots of R(X) (mod p) is 3. At least it seems so, as Q(X) (mod p) has 7 roots only for the primes 29, 43, 71 and those are primes where R(X) have three roots. But note that the converse is false, that is R(X) may have 3 roots without Q(X) having 7, for example p = 41. Now we turn to question a. : the proportion of primes for which there is a given number of roots. Looking at the number in each example seems to indicate that the proportion has a simple limit when the number of primes involved becomes bigger. It leads to the conjecture: (v) The proportion of primes p < N for which P (X) has 0 (resp. 1, 3) roots has a limit when X goes to infinity, which is 1/3 (resp. 1/2, 1/6) (vi) The proportion of primes p < N for which Q(X) has 1 roots (resp. 7) roots has a limit when X goes to infinity, which is 5/6 (resp. 1/6) (vii) The proportion of primes p < N for which R(X) has 0 roots (resp. 3) roots has a limit when X goes to infinity, which is 2/3 (resp. 1/3) Note that for R(X), there seems to exist only one prime, namely 7 for which R(X) (mod p) has exactly one root. So the proportion of such primes among the ones less than N should go to 0 when N goes to infinity. Finally, we turn to question b.. For P (X) there seem to be no rule predicting the number of roots modulo p in terms of p: it looks like pure chaos. But for Q(X) and R(X) there seem to be a simple rule determining the number of roots in terms of simply p (mod 7). (viii) If p 1 (mod 7) them Q(X) has 7 roots (mod p). Otherwise it has only one roots. (ix) If p 1 (mod 7) or p 6 (mod 7) then Q(X) has 3 roots (mod p). Otherwise it has no root, excepted when p = 7 in which case it has 1 root Theory. It turns out that all the conjectures made above are correct. It would be too hard to ask you to prove them all, but you can prove some of them. Remember first how we proved in class that a polynomial S(X) with integral coefficients monic of degree d 1 (that is of the form X d + terms of lower degrees) has at most d (incongruent) roots modulo p. Either S(X) has no root modulo p and we are done, for 0 d, or it has a root α 1 and then we can write S(X) (X α 1 )S 1 (X) (mod p), where S 1 (X) is a polynomial with integral coefficients monic of degree d 1. Now again either S 1 (X) has no root and we are done, or it has a root α 2 and we can write S(X) = (X α 1 )(X α 2 )S 2 (X) with S 2 monic of degree d 2. We can continue like that until we can t, so finally we get S(X) = (X α 1 )(X α 2 )... (X α k )S k (X) for some integer k satisfying 0 k d, and S k (X) where S k (X) is monic of degree d k and has no roots modulo p (or we do that again). Then the roots of S(X)

12 12 EXTRA CREDIT FOR MATH 39 (mod p) are α 1,..., α k, and since k d, we see that S(X) cannot have more than d roots modulo p. Here we should not jump to the conclusion that S(X) has k roots modulo p, as it could be true that some α i is congruent modulo p with some α j. If that happen, the the number of distinct roots will be less than k, and we say that S(X) has multiple root modulo p. However, it can be proved that P (X) has multiple roots only for p =.., and that Q(X) and R(X) has multiple root mod p only if p = 7. This is not very hard, but uses the notion of derivative of a polynomial that it would be too long to introduce here, so we shall admit it. 1. Prove that P (X) and R(X) have never 2 roots modulo p, and that Q(X) has never 6 roots modulo p (First assume that the polynomial has no multiple roots modulo p, and then solve the remaining cases by hand). This proves observation (i), and part of (ii) and (iii). We shall now prove observation (iii) 2. Prove that Q(X) has always at least one root modulo p, namely 1. Assume that it has more than one root module p and let z be a root of Q(X) modulo p, with z 1 (mod p). Prove that the integers 1, z, z 2, z 3, z 4, z 5 and z 6 are incongruent modulo p, and are all roots of Q(X). So Q(X) has 7 roots modulo p, and (ii) is proved. 3. Let z be as in the above exercise. By direct computation show that z + z 2 + z 4 is a root of R(X) (mod p). Also show that z 3 + z 5 + z 6 is a root of R(X) (mod p). Then prove observation (iv). So far we have prove observations (i), (ii), and (iv), but to prove observation (iii), it remains to prove that R(X) (mod p) may never have exactly one root, excepted when p = 7. You can try to do it, as a bonus question We know turn to observation (viii) and (ix) 4. Assume that Q(X) has seven roots modulo p, and let z be a root different from 1. By Fermat s little theorem, z p 1 1 (mod p). Deduce (using exercise 2) that p 1 (mod 7). To prove (viii), we would need to prove the converse of the preceding statement, that is that if p 1 (mod 7), then Q(X) has 7 roots modulo p, It is sufficient to prove that Q(X) has 1 roots different from 1, in view of exercise Assume p 1 (mod 7) and write p 1 = 7k. Let g be a primitive root modulo p. Show that z := g k is a root of Q(X) (mod p) and is not congruent to 1 modulo p. So we have proved (viii)! It was not so hard. To prove (ix) is a little bit harder, and we shall not do it, but we shall observe that a small part of it is now easy:

13 EXTRA CREDIT FOR MATH Using exercises 3 and 5, show that R(X) has 3 roots if p 1 (mod 7). Finally, it would be much too hard to prove observations (v), (vi) and (vii) from scratch, but just observe that if we admit the Theorem of Dirichlet on primes that are congruent to a modulo b (they are infinitely many if a and b are relatively prime, and the proportion of such prime less than N among all prime less than N goes to 1/φ(a) when N goes to infinity.), then one can 7. prove that observation (viii) implies observation (vi) and that observation (ix) implies observation (vii). We shall not try to prove observation (v) Further questions. Even if we have proved a significant part of the observations (i) to (ix), many questions remain. Why does the proportion of primes for which our polynomials has a given number of roots seem to converge to simple rational numbers? Is that true for all polynomials with integral coefficient or is that true just by chance for our three examples? Why is there a simple rule on p determining the number of roots of Q(X) and R(X) (mod p) (as we have seen there is one for all quadratic polynomials), but apparently not for the polynomial X 3 2? Why can the cubic polynomial P (X) = X 3 2 have exactly 1 root modulo p for infinitely many primes (actually for half of them), while the polynomial R(X), which is also cubic, can never have exactly one root modulo p (excepted for one prime p = 7, and in this case 1 is actually a multiple root)? What makes those polynomials behave so differently? The answers to those questions lie far beyond the scope of this class, but I can name the relevant theories. First, Galois theory is the modern theory of solving polynomial equations. It attaches to any polynomial a finite group, called the Galois group, which encodes the symmetries among its roots. It turns out that the Galois group for Q(X) and R(X) are commutative (the cyclic group of order 7 and 3) respectively, but that P (X) has a Galois group which is not commutative (it is the symmetric group S 3 with 6 elements). Then, we have to use Class Field Theory, a very far-reaching generalization of the quadratic reciprocity law, which shows that there are rules predicting the number of roots of a polynomial S(X) modulo p in terms of congruences of p modulo something, if, and only if, the Galois group of the polynomial S(X) is commutative. This is why we have been able to find such a rule for Q(X) and R(X) but not for P (X). Finally, there is the theorem of Cebotarev, a far-reaching generalization of Dirichlet s theorem on primes numbers, which predicts the proportion of primes p for which a given polynomial S(X) will have exactly a roots (the answer is in term of the proportion of element of a certain type in the Galois group of S(X), and work without assuming that this group is commutative). Galois theory is taught at an advanced undergraduate level (at Brandeis in math 30b = Introduction to Algebra II) but class field theory and the Cebotarev s theorem are taught only in graduate classes.