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Solutions of Congruences and Binomial Coefficients

note

This is Lab 5 from Clint by Hugh Montgomery, edited by Elad Zelingher. All rights reserved to Hugh Montgomery.

The program PolySolv allows you to specify a polynomial ff with integral coefficients, and a modulus mm, and then it evaluates f(a)(modm)f(a)(\bmod m) for each a,0a<ma, 0 \leq a<m.

Problem 1

Use PolySolv to find all roots of 7x1(mod91);7 x \equiv 1 (\bmod 91) ; all solutions of 7x35(mod91)7 x \equiv 35(\bmod 91); all solutions of 2x1(mod101)2 x \equiv 1 (\bmod 101). Note conformity with Theorem 2.17 of NZM.

Discriminants

In the next three problems, you are asked to gather data concerning the number of roots of a polynomial f(x)0(modp)f(x) \equiv 0(\bmod p), for various ff and pp, and then to formulate a conjecture. Do not be disturbed if your numerical evidence is too meager to be compelling. Each polynomial ff has a discriminant, denoted D(f)D(f), and defined on p. 487 of NZM. It can be computed using the program PolynomialDiscriminant. Prime factors of the discriminant are apt to be exceptional, and may not obey the general pattern.

Problem 2

Let f(x)=x3+x+1f(x)=x^{3}+x+1, with discriminant D(f)=31D(f)=-31. Using PolySolv, for each prime number p<100p<100 determine the value of Nf(p)N_{f}(p). What is the biggest value attained? What values are attained, and with what frequencies? What is the average of the values calculated? Formulate conjectures regarding the general situation.

Problem 3

Let g(x)=x3+x22x1g(x)=x^{3}+x^{2}-2 x-1, with discriminant D(g)=49D(g)=49. Using PolySolv, for each prime number p<100p<100 determine the value of Ng(p)N_{g}(p). What is the biggest value attained? What values are attained, and with what frequencies? What is the average of the values calculated? Formulate a conjecture regarding the general situation.

Problem 4

Let h(x)=x2+x+1h(x)=x^{2}+x+1, with discriminant D(h)=3D(h)=-3. Using PolySolv, for each prime number p<100p<100 determine the value of Nh(p)N_{h}(p). What is the biggest value attained? What values are attained, and with what frequencies? What is the average of the values calculated? Formulate a conjecture regarding the general situation.

Discussion

The situation touched on in Problems 2-4 above is quite complicated. Suppose that f(x)f(x) is a polynomial of degree dd with integral coefficients. Then 0Nf(p)d0 \leq N_{f}(p) \leq d; see Corollary 2.27 in NZM. The three polynomials considered above are irreducible (over the field Q\mathbb{Q} of rational numbers). For such polynomials, additional patterns emerge in the statistics of the Nf(p)N_{f}(p). For each k,0kdk, 0 \leq k \leq d, the primes pp for which Nf(p)=kN_{f}(p)=k have a certain relative density dkd_{k}. That is, the limit

dk=limx1π(x)pxNf(p)=k1d_{k}=\lim_{x\rightarrow\infty}\frac{1}{\pi(x)}\sum_{\substack{p\leq x \\ N_{f}(p)=k}}1

exists. These densities are determined by the Chebotarev Density Theorem in terms of the Galois group of ff. Thus the densities depend on the particular polynomial, although only finitely many configurations can arise. In the case of the polynomial of Problem 2, the Galois group is S3S_{3}, and the densities are d0=1/3,d1=1/2,d2=0,d3=1/6d_{0}=1 / 3, d_{1}=1 / 2, d_{2}=0, d_{3}=1 / 6. In Problem 3, the Galois group is C3C_{3}, and the densities are d0=2/3,d1=d2=0d_{0}=2 / 3, d_{1}=d_{2}=0, d3=1/3d_{3}=1 / 3. (For this polynomial, Nf(p)=1N_{f}(p)=1 if and only if p=7p=7.) In Problem 4 the Galois group is C2C_{2}, and the densities are d0=1/2,d1=0,d2=1/2d_{0}=1 / 2, d_{1}=0, d_{2}=1 / 2, but the situation is more elementary, since by quadratic reciprocity we find that Nf(p)=2N_{f}(p)=2 if p1(mod3)p \equiv 1 (\bmod 3), and Nf(p)=0N_{f}(p)=0 if p2(mod3)p \equiv 2 (\bmod 3). The densities then follow by the prime number theorem for arithmetic progressions.

Concerning the densities dkd_{k}, it is obvious that k=1ddk=1\sum_{k=1}^{d} d_{k}=1, and it is easy to show that dd1=0d_{d-1}=0. Not so obviously, the dkd_{k} also satisfy the relation k=1dkdk=1\sum_{k=1}^{d} k d_{k}=1. That is,

limx1π(x)pxNf(p)=1\lim_{x\rightarrow\infty}\frac{1}{\pi(x)}\sum_{p\leq x}N_{f}(p)=1

for any irreducible polynomial with integral coefficients. This is a consequence of the prime ideal theorem (which is a natural extension of the prime number theorem to algebraic number fields). For a more detailed account of how the densities dkd_{k} are calculated, see H. Heilbronn, Zeta-functions and L-functions, Algebraic Number Theory (Brighton, 1965), Thompson, Washington, 1967, pp. 204-230, especially pp. 227-229.

Problem 5

For any two polynomials f(x)f(x) and g(x)g(x), one can define their resultant, R(f,g)R(f, g). We skip the definition and fundamental theorems concerning this quantity, and mention just three useful properties: (i) If ff and gg have integral coefficients, then R(f,g)R(f, g) is an integer. (ii) R(f,g)=0R(f, g)=0 if and only if ff and gg have a common factor (i.e., a common polynomial divisor of degree >0>0 ). (iii) There exist polynomials u(x)u(x) and v(x)v(x) with integral coefficients such that

f(x)u(x)+g(x)v(x)=R(f,g).f(x) u(x)+g(x) v(x)=R(f, g) .

Suppose that ff and gg have a common root (modp)(\bmod p). That is, there is an a(modp)a(\bmod p) such that both f(a)0(modp)f(a) \equiv 0(\bmod p) and g(a)0(modp)g(a) \equiv 0 (\bmod p). On setting x=ax=a in the identity above, we see that the left hand side is divisible by pp, and hence that pR(f,g)p \mid R(f, g). Thus if pR(f,g)p \nmid R(f, g) then ff and gg have no common root, and it follows that Nfg(p)=Nf(p)+Ng(p)N_{f g}(p)=N_{f}(p)+N_{g}(p). Let ff be as in Problem 2, and gg as in Problem 3, so that f(x)g(x)=x6+x5x4+x3x2f(x) g(x)=x^{6}+x^{5}-x^{4}+x^{3}-x^{2}- 3x13 x-1. It can be shown that R(f,g)=13R(f, g)=13 in this case. By applying PolySolv, confirm that ff and gg have a common root when p=13p=13. Without performing any additional calculation, list the roots of f(x)g(x)(mod13)f(x) g(x)(\bmod 13). Apply PolySolv to fgf g, to confirm your guess.

Problem 6

Apply PolySolv with f(x)=x17321,m=1733f(x)=x^{1732}-1, m=1733. Having determined the number of roots of ff, can you deduce that mm is prime? Is this a time-effective method of proving primality? Apply PolySolv with f(x)=x17381,m=1739f(x)=x^{1738}-1, m=1739. After determining the number of roots of ff, can you deduce that mm is composite? (Recall Euler's Theorem Theorem 2.8 of NZM.) Is this a time-effective method of proving compositeness?

Problem 7

Let p=101p=101, say, and consider ff of the form f(x)=x3+ax2+bx+cf(x)=x^{3}+a x^{2}+b x+c. For various randomly-selected triples a,b,ca, b, c use PolySolv to determine the value of Nf(101)N_{f}(101). Formulate a conjecture regarding the average number of solutions of a polynomial congruence modulo a prime pp, when pp is fixed and the polynomial runs over all monic polynomials of some given degree. (A polynomial is monic if its leading coefficient is 1.) Can you prove your conjecture?

Problem 8

Let ff be defined as in Problem 2 above. Suppose that pp is a prime such that Nf(p)=2N_{f}(p)=2, and that qq is a prime such that Nf(q)=3N_{f}(q)=3 (for example p=31p = 31 and q=47q = 47). Use PolySolv to determine the value of Nf(pq)N_{f}(p q). Try some further examples of this kind. Formulate a conjecture concerning the relationship between Nf(m),Nf(n)N_{f}(m), N_{f}(n), and Nf(mn)N_{f}(m n) when gcd(m,n)=1\gcd(m, n)=1. (This conjecture is established as Theorem 2.20 in NZM, as an application of the Chinese Remainder Theorem.)

Problem 9

Suppose that pxp \nmid x. Explain why x(p1)/2±1(modp)x^{(p-1) / 2} \equiv \pm 1 (\bmod p). For how many xx does the ++ sign occur? Take f(x)=x(p1)/21f(x)=x^{(p-1) / 2}-1 in PolySolv. Try this for several values of pp. Formulate a conjecture. (This conjecture can be derived as an application of the more general Theorem 2.37 of NZM.)

Problem 10

The program PascalsT displays the entries of Pascal's Triangle (i.e., binomial coefficients), reduced (modm)(\bmod m). Start with m=2m=2. The pattern created by rows 030-3 is repeated twice in rows 474-7, with an inverted triangle of 0 's between. Does this generalize? How would you express this in terms of equations?

Problem 11

For 0n150 \leq n \leq 15, count the number of odd entries in the nn-th row of Pascal's triangle. (Take m=2m=2 in PascalsT.) The totals that arise in this way form a special class of integers. Describe it.

Problem 12

When nn is written in binary, the number of ones in the expansion is called the binary weight of nn, and is denoted w(n)w(n). That is, if n=2i1+2i2++2ikn=2^{i_{1}}+2^{i_{2}}+\cdots+2^{i_{k}} with 0i1<i2<0 \leq i_{1}<i_{2}< ik\cdots i_{k} then w(n)=kw(n)=k. Compute w(n)w(n) for 0n150 \leq n \leq 15. Note the relation between these values, and the totals computed in the preceding problem. Form a conjecture. (Problem 16 at the end of §2.2\S 2.2 of NZM is relevant here.)

Problem 13

Let pp be a prime number. What is the least nn such that p(nk)p \mid \binom{n}{k} for all kk in the range 0<k<n0<k<n ? (Take m=pm=p in PascalsT, and look for zeros.) What is the second such nn ? The third? (Problem 14 at the end of §2.2\S 2.2 of NZM is relevant here.)

Problem 14

For what k,0k15k, 0 \leq k \leq 15, is it true that 3(15k)3 \nmid\binom{15}{k} ? For what k,0k15k, 0 \leq k \leq 15, is it true that 5×(15k)5 \times\binom{ 15}{k} ? Does this suggest something?

Problem 15

Let pp be a prime number. Describe all the patterns that you can find in the sequence of residues (np)(modp2)\binom{n}{p}\left(\bmod p^{2}\right).