Homework Statement
The idea of this problem is to investigate the solutions to x^2=1 (mod pq), where p,q are distinct odd primes.
(a) Show that if p is an odd prime, then there are exactly two solutions (mod p) to x^2=1 (mod p). (Hint: difference of two squares)
(b) Find all pairs...
Homework Statement
Decipher the following text
KQEREJEBCPPCJCRKIEACUZBKRVPKRBCIBQCARBJCVFCUPKRIOF KPACUZQEPBKRXPEIIEABDKPBCPFCDCCAFIEABDKPBCPFEQPKAZ
BKRHAIBKAPCCIBURCCDKDCCJCIDFUIXPAFFERBICZDFKABICBB
ENEFCUPJCVKABPCYDCCDPKBCOCPERKIVKSCPICBRKIJPKABI
Homework Equations
I know that...
Homework Statement
Below is an example of ciphertext obtained from an Affine Cipher. Determine the plaintext.
KQEREJEBCPPCJCRKIEACUZBKRVPKRBCIBQCARBJCVFCUPKRIOFKPACUZQEPBKRXPEIIEABDKPBCPFCDCCAFIEABDKPBCPFEQPKAZ
BKRHAIBKAPCCIBURCCDKDCCJCIDFUIXPAFFERBICZDFKABICBB...
I figured out that for a its if and only if the number of digits is a multiple of d, where d divides b-1.
For b, would it just be if and only if the alternating sum comes out to a multiple of d, where d divides b+1?
Well for b=1 then \alpha will be positive so \beta will have to be the negative of \alpha
The opposite is true for b=-1, \alpha will be negative and \beta is going to be the positive of that number...
Homework Statement
A base b repunit is an integer with base b expansion containing all 1's.
a) Determine which base b repunits are divisible by factors b-1
b) Determine which base b repunits are divisible by factors b+1
Homework Equations
R_{n}=\frac{b^{n}-1}{b-1}
The Attempt...
For 3 part a, additive order of a modulo n is defined to be the smallest positive integer m that satisfies the congruence equation m*a \cong 0 (mod n). So in this case it'd be better to write a modulo n as m*a \cong 0 (mod n). m would be our additive order which means since n=78 our m=78/a?
So for question 1. [0] can occur because 2^{2}+2^{2}= 0 mod 4.
[1] can occur because 2^{2}+1^{2}= 1 mod 4.
Is [3] the only one that can not occur?
As for question 2a, I went through and squared all numbers from 1 to 20, the only options...
Homework Statement
1) What are the possible values of m^{2} + n^{2} modulo 4?
2) Let d_{1}(n) denote the last digit of n (the units digit)
a) What are the possible values of d_{1}(n^{2})?
b) If d_{1}(n^{2})=d_{1}(m^{2}), how are d_{1}(n) and d_{1}(m) related?
3) a)...
Homework Statement
Determine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the first differential equations given in #7.
Homework Equations
#7 (sin \theta)y^{'''} - (cos \theta)y^{'} = 2
#10 (u)dv +...
Homework Statement
Let A be the adjacency matrix for the graph G=(V,E)
(a) Show that A^3[i,i] equals twice the number of triangles containing vertex i. (A triangle is a cycle of length 3
(b) Find an interpretation for A^3[i,j] when i does not equal j similar to the above. Prove that...
Homework Statement
Prove that a graph is a tree if and only if it has no cycles and the insertion of any new edge always creates exactly one cycle.
The Attempt at a Solution
Assume that a graph G is connected and contains no vertices with a degree of zero.
So would I get my proof by...
Ok, so after getting the diagram, if I use that theorem to show G is Eulerian it is as easy as saying since G is a connected graph where the 2-element vertices have a degree of 2 while the 3-element vertices have a degree of 4 so they all have an even number of vertices making it Eularian...
Ok, thats what I figured but I how would I find |A and B|? Is there some kind of ordered pair subtraction I am forgetting about or something?
For example if I take A = {1,2} and B = {1,3} how do I find what |A and B| is? Does the method also work if I was to replace B with a 3-element set?
Homework Statement
Let X = {1,2,3,4} and let G = (V,E) be the graph whose vertices are the 2-element and 3-element subsets of X and where A is adjacent to B if |A and B| = 2. That is:
V = nCr(X,2) or nCr(X,3)
E = {{A,B}:A,B\inV and |A\capB|=2}
(a) Draw the diagram of the graph G...
Alright well I found out a doubling construction is just the fact that if an STS of order v, STS(v) exists then so does an STS(2v+1). So that means that my STS(19) is the same as a STS(39). Not sure what to do from there though...