18.781 Homework 1
Due: 9th February 2014
Q 1 (1.2(6). Prove that the product of three consecutive integers is divisible by 6; of four consecutive integers
by 24.
Proof. Let the integers be n, n + 1 and n + 2. If n is odd, then n = 2k + 1 for some integer
18.781: SECOND PRACTICE HOUR EXAM #1
(for Hour Exam of March 15, 2013)
This exam is closed book. Calculators are also not permitted. Do all four
problems. Be sure to show your work. Do not simply state the answer.
There are a total of 40 points.
Warning:
18.781: SOLUTIONS TO SECOND PRACTICE
HOUR EXAM #1
(for hour exam of March 15, 203)
1. One way to do this is to write a = 20x + 7 and solve 20x + 7
8 (mod 43) using the Euclidean algorithm. We get x 28 (mod 43) and
a 567 (mod 20 43).
2. By Theorem 2.15 an
18.781: PRACTICE HOUR EXAM #1
(for hour exam of March 15, 2013)
This exam is closed book. Calculators are also not permitted. Do all four
problems. Be sure to show your work. Do not simply state the answer.
There are a total of 40 points.
Warning: This pr
18.781: HOUR EXAM #1
(March 15, 2013)
This exam is closed book. Calculators are also not permitted. Do all four
problems. Be sure to show your work. Do not simply state the answer.
There are a total of 40 points.
1. (a) (5 points) Let p be prime, p > 5. S
18.781: SOLUTIONS TO HOUR EXAM #1
(for hour exam of March 15, 2013)
1. (a) Write p2 p = (p 1)p, the base p expansion of p2 p. Since
p > 5, the base p expansion of 5p is just 5p. Hence by Lucas
theorem,
(p 1)p
5p
p1
(mod p)
5
(p 1)(p 2)(p 3)(p 4)(p 5)
(mod
18.781 Solutions to Problem Set 4, Part 2
1. (a) To nd a primitive root mod 23, we use trial and error. Since (23) = 22, for a to be a primitive
root we just need to check that a2 1 (mod 23) and a11 1 (mod 23).
211 = 25 25 2 9 9 2 11 2 1
(mod 23),
311 33
18.781: SOLUTIONS TO PROBLEM SET #2
(due February 19, 2013)
1.3(4) Let n =
ai 10i be the base 10 expansion of n, so 0 ai 9. The
sum of the digits of n is
ai . We have
n
ai =
ai (10i 1).
Now 10i 1 is divisible by 9 [why?] and hence also by 3, and the proof
18.781: SOLUTIONS TO PROBLEM SET #3
(due February 25, 2013)
2.1(17) By Wilsons theorem, we know that 70! + 1 0 (mod 71). Now
(70! + 1) (63! + 1) = 63! (70 69 64 1).
Since (63!, 71) = 1, we need to show that
70 69 64 1 (mod 71).
Now
70 69 64 (1)(2) (7) (mo
18.781, Fall 2007 Problem Set 7 Due: FRIDAY, October 26
We are beginning a new section on algebraic number theory. Supporting discussion for this material can be found in Chapter 9 of NZM (though it is somewhat less focused than our in-class discussi
18.781: SOLUTIONS TO PROBLEM SET #11
(due April 29, 2013)
6.2(4) Suppose to the contrary that a/b were such a rational number. Then
with the unique exception h = a , we have
k
b
ah
1
.
bk
bk
1
Hence if a h < k, then bk < k so k 1 < 1 . Since > 1, this
b
k
18.781: SOLUTIONS TO PROBLEM SET #10
(due April 22, 2013)
5.4(4) Let 2a x, 2b y , and 2c z . Let u = x/2a , v = y/2b, and w = z/2c , so
u, v , w are odd. We may suppose by symmetry that a b c. If
x2 + y 2 + z 2 = 2xyz , then
u2 + 22(ba) v 2 + 22(ca) w 2 =
18.781: SOLUTIONS TO PROBLEM SET #8
(due April 8, 2013)
(A10) (1) This is a restatement of Theorem 2.20 on page 70.
(2) Let (m, n) = 1. Recall that if f (x) is a polynomial with integer
coecient and a Z, then
f (a + m) f (a) (mod m).
(5)
Now let 1 j m and
18.781: SOLUTIONS TO PROBLEM SET #7
(due April 1, 2013)
3.1(6) (a) Answer in back of book. To do the computation, simply com2
pute 12 , 22 , . . . , p1 and reduce modulo p. (See the rst part of
2
Problem 14.)
(b) Fix n. Then n2 is a quadratic residue modu
18.781: SOLUTIONS TO PROBLEM SET #6
(due March 18, 2013)
(A7) We have 8200 = (23 )200 = 2600 1 (mod 601) by Fermats theorem.
Hence ord601 (8) 200 < 600, so 8 is not a primitive root. In general, if
d|(p 1) and d > 1, then any number of the form ad is not
18.781: SOLUTIONS TO PROBLEM SET #5
(due March 11, 2013)
2.7(2) Divide x7 x by 2x3 + 5x2 + 6x + 1 using the ordinary long division
algorithm for polynomials, but regarding the coecients modulo 7.
(You should know how to do this kind of division, so try it
18.781: SOLUTIONS TO PROBLEM SET #4
(due March 4, 2013)
2.3(4) By the Chinese remainder theorem, there will be a unique solution x
modulo 3 4 5 = 60. Trial and error shows that x 58 (mod 60). A bit
more ecient way to proceed is to rst use that x 3 (mod 5)
18.781 Solutions to Problem Set 5
1. Note that 41 1 = 23 5. Start with a quadratic nonresidue mod 41, say, 3. Now b = 35 = 81 3 3
(mod 41), which has order exactly 8. (3)1 14 (mod 41).
Now we calculate a square root of 21. First, check that 21 is a square
18.781 Solutions to Problem Set 2
k
+
1. Let m = n k. We want to show that the power of p dividing mk k = (m+k!)! is the number of carries
m!
when adding m to k in base p. Note that each time a carry occurs, (ai + p) in the ith place becomes
ai in the ith
18.781 Homework 6
Due: 18th March 2014
Q 1 (2.8(2). Find a primitive root of 23.
Proof. We can check that 5 is a primitive root for the prime 23. (23) = 22, so 522 1(mod 23). But
52 = 25 1(mod 23) and 511 (52 )5 5 25 5 9 5 1(mod 23) 1(mod 23).
Q 2 (2.8(5)
18.781 Introduction to Number Theory
Prof. A. De Sole
Mid Term Exam1
March 11, 2014
First and Last Name:
MIT ID number:
Problem #
Total Points
1
10
2
10
3
10
4
10
5
10
Total
50
Justify your answers!
1
Score
Problem 1. Check that 259 is invertible modulo 1
18.781 Homework 5
Due: 11th March 2014
Q 1 (2.6(2). Solve x5 + x4 + 1 0(mod 34 ).
Proof. 1(mod 3) is the only solution modulo 3 and the derivative 5x4 +1 vanishes at 1(mod 3), but 15 +14 +1
0(mod 9), so there are no solutions to this equation mod 9 and t
18.781 Homework 3
Due: 25th February 2014
Q 1 (2.1(26). Show that the product of three consecutive integers is divisible by 504 if the middle one is a
cube.
Proof. 504 = 23 32 7. Let the three consecutive numbers be cfw_x3 1, x3 , x3 + 1. Their product is
18.781 Introduction to Number Theory
Prof. A. De Sole
Mid Term Exam1
March 11, 2014
First and Last Name:
Alberto De Sole
MIT ID number:
1234567890
Problem #
Total Points
1
10
10
2
10
10
3
10
10
4
10
10
5
10
10
Total
50
50
Justify your answers!
1
Score
Pro
18.781 Homework 2
Due: 18th February 2014
Q 1 (1.3(3). Prove that a number is divisible by 2 if and only if the units digit is divisible by 2; that a
number is divisible by 4 if and only if the integer formed by its tens and units digit is divisible by 4;
18.781 Homework 4
Due: 4th March 2014
Q 1 (2.2(5)(e-g). Find all solutions to the equations (e): 64x 83(mod 105), (f ): 589x 209(mod 817),
(g): 49x 5000(mod 999).
Proof.
(e)
By running the Euclidean algorithm, we get 25 105 41 64 = 1, so the unique soluti
18.781 Solutions to Problem Set 3
1. Its enough to solve the congruence mod 11 and mod 13, and then combine the solutions by Chinese
Remainder Theorem. Now x3 9x2 + 23x 15 factors as (x 1)(x 3)(x 5), so solutions mod 11 or
mod 13 are 1, 3, 5 in each case.