Number Theory Midterm
Study Guide
Review Session – Monday March 1 in IT 222
Some key definitions/Theorems to know:
a. Well Ordering Principle (We assumed this, we never proved it. It’s the assumption the
whole course comes from.)
b. First and Second Principles of Finite Induction
c. Pascal’s Rule
d. Binomial Coefficient
e. Binomial Theorem
f. Triangular Number
g. Division Algorithm
h. Greatest Common Divisor
i. Relatively Prime
j. Euclid’s Algorithm
In addition, you should know theorems 2.2-2.5 Note:
2.4 tells us a and b are relatively prime iff there exist integers x and y with ax+by=1.
2.5 says that if a|bc with a and b relatively prime, then a|c
You do not need to study 2.6 for the midterm.
You should be able to apply the Euclidean Algorithm to do both:
a. Given integers a and b, find their gcd
b. Given integers a and b, express the gcd of a and b as a linear combination.
You should also recall from page 18 that the square of any integer is of the form 4k or 4k+1
for some k and that the square of any odd integer is of the form 8j+1 for some j. (You
should also be able to prove these, but unless asked to prove, you can assume them. )
Some example problems: (Two of the test questions will come from these. I expect to add
one more problem on Monday to this list.)
1. Show that 2.6.10… (4n-2)=(2n)!/n!
** Proof by induction
For the basis, let n=1. The left side is 2. The right side is (2x1)!/1!=2
For the induction step, assume the statement is true for n=k. So:
2.6….(4k-2)=(2k)!/k!
Then for n=k+1, the left side:
2  6 ... (4k  2)(4(k 1)  2) 
2k !
k!
4k  2
by the induction hypothesis. The right side is:
2k 1!  2k  22k 12k !  22k 12k !  4k  22k !
k 1!


(k 1)k!
k!
k!
2. More than any other method, we’ve used the well ordering principle to prove theorems.
Don’t complete the proof, but explain how we used the WOP to set up the proof that given 2
integers, not both 0, there exists x and y such that gcd(a,b)=ax+by (Theorem 2.3)
** We begin by defining the set S as follows:
S  au bv | u,v  I,au bv  0Then we prove S is not the empty set. The W.O.P.
tells us that there is a least element in S. We then show that this element divides both a and
b using the fact that it is the least element of S .

3. Show that the cube of any integer has one of the forms 9k, 9k+1 or 9k-1 for some k.
**From the DA, every integer can be written as 3k or 3k+1 or 3k-1 (any 3k-1=3(k-1)+2)
The cubes of these are 27k, 27k3+3(27k2)+3(3k)+1=9m+1,
and 27k3-3(27k2)+3(3k)-1=9n-1 for appropriate m or n.
4. Expand (a+b)7
7
7
**

(a  b)  a  7a6b  21a5b2  35a 4 b 3  35a 3b 4  21a2b5  7ab6  b 7
5. Find the gcd(1617,525)
**Use EA:
1617= 3x525+42
525=12x42+21
42=2x21+0
gcd(1617,525)=21
6. Find x and y so the gcd(24,138)=24x+138y using the Euclidean Algorithm
**use EA to find gcd
138=5x24+18
24=1x18+6
18=3x6+0 so gcd=6
Then 6=24-1x18
=24-1x(138-5x24) = 6x24-1x138
7. For any integer x, show a|(ax+b) iff a|b
**Part 1. Assume a|ax+b Then there exists m such that am=ax+b So am-ax=b
So a(m-x)=b So a|b
Part 2. Assume a|b Then there exists some integer m such that am = b Then
ax+b = ax+am = a(x+m) So a|ax+b
8. Show that if a|bc then a| gcd(a,b)gcd(a,c)
** Let d=gcd(a,b) and e=gcd(a,c). Then by Theorem 2.3, there exist integers x,y,m, and n
such that d=ax+by and e=am+cn.
Then gcd(a,b)gcd(a,c)=d.e=(ax+by)(am+cn)=a2xm+axcn+abmy+bcny. Since a|bc, a divides
every term of this expression so a| gcd(a,b)gcd(a,c)
9. (Added Monday night)
The nth Catalan number for n  0, is defined to be
Prove
Cn 
22n 1
Cn1 for n1.
n 1
Cn 
1 2n 
 
n 1  n 

** Proof:
Calculate out what the left side is and what the right side is using the definition above.

1 2n  1 2n!
2n!

Left side is Cn 
 
n 1  n  n 1 n!n! n 1!n!
Right side is
22n 1
22n 1 1 2(n 1) 22n 1 1 2n  2!
Cn1 



n 1
n 1 n  n 1 
n 1 n n 1! n 1!

22n 1!
22n 1! n
2n!


n 1! n 1! n 1! n 1! n n 1! n!
Note the second last step involves multiplying both the numerator and denominator by n.
