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- From: goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard)
- Subject: More number theory problems (complete)
- Message-ID: <1992Sep5.144920.5797@cs.rose-hulman.edu>
- Sender: news@cs.rose-hulman.edu (The News Administrator)
- Nntp-Posting-Host: g214-1.nextwork.rose-hulman.edu
- Organization: Rose-Hulman Institute of Technology
- Date: Sat, 5 Sep 1992 14:49:20 GMT
- Lines: 159
-
-
- Thanks to those who have sent solutions. This is a long post,
- wherein I update the problem list. Last time I referred to some
- problems only by number. This time, "everything is here". Please
- email solutions. To refresh your memory:
-
- >So here's the deal. If you send me a solution to one of the problems
- >below, and I like, it I might send you 2 bucks. Also, I'll post nice
- >solutions for the benefit of your ego!
-
- 5.2.6 Is solved. It had a typo (change the last 2 to a 1).
-
- 5.3.4 is solved. (It was trivial, <blush>.)
-
- 2.2.24 Let a and b be positive integers and let r_j and q_j be the
- remainders and quotients of the steps of the Euclidean algorithm as
- defined in "this section".
- a) find the value of \sum_{j=1}^{n} r_j q_j.
-
- b) find the value of \sum_{j=1}^{n} (r_j)^{2} q_j.
-
- 8.6.10 is solved.
-
- 8.6.14 Find all Charmicheal numbers of the form 5pq where p and q are
- primes. (I have an ugly solution, maybe the casework could be cut
- down?)
-
- 8.6.18 Show that if (a,n) = 1 then q_n(a+nc)==
- q_n(a + \lambda(n) c a^{-1} (mod n). (We define q_n(a) ==
- (a^{\lambda(n)}-1/n modulo n, with 0 <= q_n(a) < n.)
- (\lambda(n) is the minimal universal exponent.)
-
- HERE'S THAT DOUBLE STARRED PROBLEM YOU'VE BEEN ASKING FOR!:
- 9.4.8 Let the composite positive integer n have prime factorization
- n= p_{1}^{a_1}...p_{m}^{a_m}, where p_j = 1+2^{k_j}q_{j} for j = 1, 2,
- .. m, where k_1 <= k_2 <= ... <= k_m, and where n=1+2^{k}q. Show that
- n is an Euler pseudoprime to exactly
-
- \delta_n \prod_{j=1}^{m} gcd((n-1)/2, p_{j}-1)
-
- different bases b, with 1<= b < n , where
-
- ( 2 if k_1 = k
- (
- \delta_n = (1/2 if k_j < k and a_j is odd for some j
- (
- ( 1 otherwise.
-
- (An Euler pseudoprime to the base b in an odd composite which satisfies
-
- b^{(n-1)/2} == (b/n) (mod n)
-
- (b/n) is the Jacobi symbol.
-
- Whew! That's probably a result from someones paper. A reference would
- be sufficient for my needs.
-
- 10.4.18 Is solved. It should have read "D = (ta_k +1)^2 +
- 2ta_{k-1} +1".
-
- A PREAMBLE:
- The greatest common divisor of two integers can be found using only
- subtractions, parity checks, and shifts of binary expansions, without
- using any divisions. The algorithm proceeds recursively using the
- following reduction:
-
- ( a if a=b
- (
- ( 2(a/2,b/2) if a and b are even
- (a,b) = (
- ( (a/2,b) if a is even and b is odd
- (
- ( (a-b,b) if a and b are odd and a>b.
-
- Reverse the roles of a and b when necessary.
-
- 2.2.12 Show that to find (a,b) this algorithm uses the subtraction
- step in the reduction no more than 1 + [log_{2} ( max(a,b) )] times.
-
- ANOTHER PREAMBLE:
- Briefly, we want to consider the "least-remainder algorithm". It is
- analogous to the Euclidean algorithm, but each remainder is chosen
- to have the smallest absolute value instead of the smallest nonnegative
- value. Let r_0 = a and r_1 = b. Then we have
-
- r_0 = r_1 q_1 + e_2 r_2, -r_1/2 < e_2 r_2 <= r_1/2
- .
- .
- .
- r_{n-2} = r_{n-1} q_{n-1} + e_n r_n, -r_{n-1}/2 < e_n r_n <=
- r_{n-1}/2
- r_{n-1} = r_n q_n
-
- and (a,b) = r_n. Note that r_j is always positive; we let e_j = +-1 be
- the sign.
-
- 2.2.16 (double starred!) Show that the least-remainder algorithm is
- always faster or as fast as the Euclidean algorithm. (Hint: First show
- that if a and b are positive integers with 2b<a, than the least
- remainder algorithm can find (a,b) in no more steps than it uses to
- find (a, a-b).)
-
- 2.2.18 Show that the number of divisions needed to find the gcd of two
- positive integers using the least remainder algorithm is less than
- 8/3 times the number of digits in the smaller of the two numbers, plus
- 4/3.
-
- YET ANOTHER PREAMBLE:
-
- The game of Euclid: Two players begin with a pair of positive integers
- and take turns making moves of the following type. A player can move
- from the pair {x,y}, with x>=y, to any of the pairs {x-ty,y}, where t
- is a positive integer and x-ty >=0. A winning move consists of moving
- to a pair with one element equal to 0.
-
- For reference, 2.2.21 (which is solved) states: Every sequence of moves
- starting at {a,b}, must end with {0,(a,b)}.
-
- 2.2.22 Show that ina game beginning with the pair {a,b}, the first
- player may play a winning strategy if a=b or if a > b(1 + \sqrt{5})/2;
- otherwise, the second player may play a winning strategy. (Hint:
- First show that if y<x<=y(1 + \sqrt{5})/2 then there is a unique move
- from {x,y} that goes to a pair {z,y} with y>z(1 + \sqrt{5})/2.)
-
- In the next problem, part (a) is solved but part (b) is stubborn. It's
- not a starred problem, so I'm starting to suspect a typo...
-
- 5.2.11.a (Solved) Show that if n is a pseudoprime to the base a but not
- a pseudoprime to the base b, where (a,n)=(b,n)=1, then n is not a
- pseudoprime to the base ab.
-
- 5.2.11.b Show that if there is an integer b with (b,n)=1 such that n is
- not a pseudoprime to the base b, then n is a pseudoprime to <= \phi(n)
- different bases a, with 1<=a<n. (Hint: Show that the sets a_1, a_2,
- .., a_r, and ba_1, ba_2,...,ba_r have no common elements, where a_1,
- a_2, ..., a_r, are the bases less than n to which n is a pseudoprime.)
-
- 8.6.16 (Should be easy if one knows all the notation and terminology)
- Show that the deciphering exponent d for an RSA cipher with enciphering
- key (e,n) can be taken to be an inverse of e mod \lambda(n) (minimal
- universal exponent). (n = pq is the modulus, and e in the enciphering
- exponent.)
-
- THE NEXT TWO PROBLEMS ARE FROM THE "PSEUDO-RANDOM NUMBER" SECTION
-
- 8.7.6 Show that every linear congruential pseudo-random number
- generator can be simply expressed in terms of a linear congruential
- generator with increment c=1 and seed 0, by showing that the terms
- generated by the linear congruential generator x_{n+1} == ax_{n} + c
- (mod m), with seed x_0, can be expressed as x_n == b y_{n} + x_0 (mod
- m), where b == (a-1)x_0 + c (mod m), y_0 = 0 and y_{n+1} == ay_n +1
- (mod m).
-
- 8.7.8 Show that the maximal possible period length for a pure
- multiplicative generator of the form x_{n+1} == ax_n (mod 2^e), e>=3,
- is 2^{e-2}. Show that this is obtained when a == +-3 (mod 8).
-
- Bart Goddard
- goddard@nextwork.rose-hulman.edu
-