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Question - Math 324 Assignment 7 (due November 5)

1. (11.1 # 12 b)) Consider the quadratic congruence ax2 + bx + c ≡ 0 mod p, where p is an
odd prime and a, b, c are integers with (a, p) = 1. Let d = b2 − 4ac, and show that
a) the congruence ax2 + bx + c ≡ 0 mod p is equivalent to y2 ≡ d mod p, with y = 2ax + b.
b) if d ≡ 0 mod p there is exactly one solution x mod p; if d is a quadratic residue there
are exactly two solutions mod p; and if d is a quadratic non-residue there is no solution.

2. Show that ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
p
2 = 1 if p ≡ 1, 3 mod 8, and = − 1 if p ≡ − 1, − 3 mod 8.

3. (11.1 #14). Show that, if p ≥ 7 is prime, then ...Read More there are always two consecutive
quadratic residues mod p. Hint: First show that at least one of 2, 5, and 10 is a quadratic
residue mod p.

4. (4.4 # 10) How many solutions mod 144 are there to the congruence x5 + x − 6 ≡ 0
mod 144? Explain why.

5. For all k ≥ 3, show that
a) 5 has order 2k − 2 mod 2k. Hint: First, show that 5 raised to the power 2k − 3 equals
1 + 2 k − 1xk with xk odd, by induction on k ≥ 3.
b) each a with (a, 2k) = 1 has a ≡ (− 1)α5β mod 2k, for unique α, β with 0 ≤ α < 2 and
0 ≤ β < 2 k − 2. Hint: There are φ(2k) = 2⋅2k − 2 such a mod p. How many (− 1)α5β mod p?

6. Eisenstein’s Criterion: The polynomial f(x) = x n + a n − 1x n − 1 + … + a1x + a0,with
integer coefficients and n ≥ 1, is irreducible over Q if there exists a prime number q
dividing a0, a1, …, a n − 1 so that q2 does not divide a0.
Assuming Eisenstein’s criterion, prove that the qth cyclotomic polynomial
Φ(x) = (xq − 1)/(x − 1) = x q − 1 + x q − 2 + …+ x + 1
is irreducible over Q, as follows:
(a) Show that Φ(x) is irreducible if, and only if, Φ(x + 1) is irreducible.
(b) Show that Φ(x + 1) = x q − 1 + x⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−1q
q q − 2 + …+ x + . ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
2
q
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
1
q
(c) Apply Eisenstein’s criterion to show that Φ(x + 1), and hence Φ(x), is irreducible. ...Read Less

Solution Preview - c and y = 2ax + b, we get an identity y2 − d = 4a(ax2 + bx + c). Since (4a, p) = 1, it follows that x solves ax2 + bx + c ≡ 0 mod p ⇔ y solves y2 ≡ d mod p. Also y = 2ax + b, with (2a, p) =1, implies that the number of solutions x mod p is the same as the number of solutions y mod p. Remark: This is just ‘completing the square’. b)

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