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Congruence - Number Theory - Solved Exam, Exams of Number Theory

This is the Solved Exam of Number Theory which includes Last Complete Solution, Prime, Integer Relatively, Root Modulo, Distinct Primes, Primitive Root Modulo, Discrete Logarithms, Incongruent etc. Key important points are:Congruence, Prime Number, Two Congruences, Integer, Satis Es, Prime Number, Seven Binomial CoeCients, Numbers, Non Negative Integer, Square Roots

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Math 115 Professor K. A. Ribet
First Midterm Exam February 25, 1998
Answer question #2 and three other questions.
1(6 points).Find all solutions to the congruence x2pmod p2when pis a prime number.
There are no solutions. Indeed, if xsatisfies the congruence, then x20 mod p. Thus pdivides x2, so
pdivides xand thus p2divides x2. Since we then have x20 mod p2, we do not have x2pmod p2.
(Hensel’s lemma has nothing to do with this problem: it doesn’t apply, so it gives no information.)
2(9 points).Using the equation 7·529 3·1234 = 1, find an integer xwhich satisfies the two congruences
xn123 mod 529
321 mod 1234 and an integer ysuch that 7y1mod 1234. (No need to simplify.)
This is the question that you were more or less promised. I would take xto b e 123 ·(3) ·1234 + 321 ·7·529,
or 733317; you can reduce this mod 652786, getting 80531 instead. Take yto be 529.
3(7 points).Suppose that pis a prime number. Which of the p+ 2 numbers p+ 1
k(0kp+ 1) are
divisible by p?
This was a “frequently omitted” question, but it’s reasonably easy. First proof: The formula p+ 1
k=
(p+ 1)!
k!(p+ 1 k)! displays p+ 1
kas a fraction whose numerator is divisible by pbut not by p2and whose
denominator is divisible by ponly when kor p+ 1 kis one of the two numbers p,p+ 1. Hence the
coefficient is divisible by pfor all kexcept the values 0, 1, p,p+ 1. Among the p+ 2 numbers in question,
p2 of them are divisible by p. Second proof: Think about Pascal’s triangle and the fact that all the
numbers in the pth row are divisible by p(except for the two 1’s at the ends).
4(7 points).Let pbe a prime and let nbe a non-negative integer. Suppose that ais an integer prime to p.
Show that b:= apnsatisfies bamod pand bp11mod pn+1.
If ais as in the problem, then ap11 mod pby Fermat’s little theorem. Thus ais a root mod pof
f(x) = xp11. Since f0(a) is then prime to p,acan be refined in exactly one way to a root of f(x)
mod pn+1. In other words, there exists a unique bmod pn+1 which satisfies bp11 mod pn+1 and ba
mod p. The p oint of this problem is that you can actually exhibit a bthat works. Namely, using Fermat’s
congruence apamod p, you easily prove apn
amod pby induction. Thus bis indeed the same as a
mod p. Also, ais prime to pn+1, so aφ(pn+1 )1 mod pn+1. Since φ(pn+1 ) = (p1)pn, this yields bp11
mod pn+1.
5(6 points).Show that n4+n2+ 1 is composite for all n2.
I added this problem at the last minute, just taking it from the book (problem 33 on page 32). If you calculate
two or three of these numbers, you see that they are composite but are not systematically divisible by any
particular number. This suggests an algebraic factorization. The point turns out to be that n4+n2+ 1 =
(n4+ 2n2+ 1) n2is a difference of two squares. It’s thus the product n2+1+nn2+ 1 n. To prove
that this is a non-trivial factorization, you have to see that n2n+1 is bigger than 1, but this is easy (graph
it, or something).

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Math 115 Professor K. A. Ribet

First Midterm Exam February 25, 1998

+ Answer question #2 and three other questions.

1 (6 points). Find all solutions to the congruence x^2 ≡ p mod p^2 when p is a prime number.

There are no solutions. Indeed, if x satisfies the congruence, then x^2 ≡ 0 mod p. Thus p divides x^2 , so p divides x and thus p^2 divides x^2. Since we then have x^2 ≡ 0 mod p^2 , we do not have x^2 ≡ p mod p^2. (Hensel’s lemma has nothing to do with this problem: it doesn’t apply, so it gives no information.)

2 (9 points). Using the equation 7 · 529 − 3 · 1234 = 1, find an integer x which satisfies the two congruences

x ≡

123 mod 529 321 mod 1234

and an integer y such that 7 y ≡ 1 mod 1234. (No need to simplify.)

This is the question that you were more or less promised. I would take x to be 123 · (−3) · 1234 + 321 · 7 · 529, or 733317; you can reduce this mod 652786, getting 80531 instead. Take y to be 529.

3 (7 points). Suppose that p is a prime number. Which of the p + 2 numbers

(p + 1 k

( 0 ≤ k ≤ p + 1) are divisible by p?

This was a “frequently omitted” question, but it’s reasonably easy. First proof: The formula

p + 1 k

(p + 1)! k!(p + 1 − k)!

displays

(p + 1 k

as a fraction whose numerator is divisible by p but not by p^2 and whose

denominator is divisible by p only when k or p + 1 − k is one of the two numbers p, p + 1. Hence the coefficient is divisible by p for all k except the values 0, 1, p, p + 1. Among the p + 2 numbers in question, p − 2 of them are divisible by p. Second proof: Think about Pascal’s triangle and the fact that all the numbers in the pth row are divisible by p (except for the two 1’s at the ends).

4 (7 points). Let p be a prime and let n be a non-negative integer. Suppose that a is an integer prime to p. Show that b := ap

n satisfies b ≡ a mod p and bp−^1 ≡ 1 mod pn+1.

If a is as in the problem, then ap−^1 ≡ 1 mod p by Fermat’s little theorem. Thus a is a root mod p of f (x) = xp−^1 − 1. Since f ′(a) is then prime to p, a can be refined in exactly one way to a root of f (x) mod pn+1. In other words, there exists a unique b mod pn+1^ which satisfies bp−^1 ≡ 1 mod pn+1^ and b ≡ a mod p. The point of this problem is that you can actually exhibit a b that works. Namely, using Fermat’s congruence ap^ ≡ a mod p, you easily prove ap

n ≡ a mod p by induction. Thus b is indeed the same as a

mod p. Also, a is prime to pn+1, so aφ(p

n+1) ≡ 1 mod pn+1. Since φ(pn+1) = (p − 1)pn, this yields bp−^1 ≡ 1 mod pn+1.

5 (6 points). Show that n^4 + n^2 + 1 is composite for all n ≥ 2.

I added this problem at the last minute, just taking it from the book (problem 33 on page 32). If you calculate two or three of these numbers, you see that they are composite but are not systematically divisible by any particular number. This suggests an algebraic factorization. The point turns out to be that n^4 + n^2 + 1 = (n^4 + 2n^2 + 1) − n^2 is a difference of two squares. It’s thus the product

n^2 + 1 + n

n^2 + 1 − n

. To prove

that this is a non-trivial factorization, you have to see that n^2 − n + 1 is bigger than 1, but this is easy (graph it, or something).