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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, SatisEs, Prime Number, Seven Binomial CoeCients, Numbers, Non Negative Integer, Square Roots
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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).