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 6≡
0(mod 9), so there are no solutions to this equation mod 9 and therefore none modulo 34 .
Q 2 (2.6(3)). Solve x3 + x + 57 ≡ 0(mod 53 ).
Proof. 43 + 4 + 57 = 125 ≡ 0(mod 53 ) and the derivative of this polynomial 3x2 + 1 is non-zero at 4(mod 5),
so 4(mod 125) is the unique solution modulo 125.
Q 3 (2.6(4)). Solve x2 + 5x + 24 ≡ 0(mod 36).
Proof. We have to solve the equation x2 + 5x + 24 = 0 modulo 4 and 9 and use Theorem 2.20 to construct
solutions modulo 36. Modulo 4, it is easy to see that 0(mod 4) and 3(mod 4) are the only solutions.
Modulo 3 we get the two solutions 0(mod 3) and 1(mod 3). The derivative 2x + 5 is non-zero at both
these roots, so they lift unique to roots modulo 9. Looking through the sets {3a(mod 9)|0 ≤ 1 ≤ 2}
and {1 + 3b(mod 9)|0 ≤ 1 ≤ 2} we see that 6(mod 9) and 7(mod 9) are the unique solutions modulo
9. 1(mod 4) is the inverse to 9(mod 4) and 7(mod 9) is the inverse to 4(mod 9). This tells us that the
solutions modulo 36 are the residue classes of 28 ∗ 6, 28 ∗ 7, 28 ∗ 6 + 9 ∗ 3, 28 ∗ 7 + 9 ∗ 3. Simplifying this gives
{7(mod 36), 15(mod 36), 16(mod 36), 24(mod 36)}.
Q 4 (2.6(9)). Suppose that f (a) ≡ 0(mod pj ) and that f 0 (a) 6≡ 0(mod p). Let f 0 (a) be an integer chosen so
that f 0 (a)f 0 (a) ≡ 1(mod pj ), and put b = a − f (a)f 0 (a). Show that f (b) ≡ 0(mod p2j ).
Proof. The Taylor expansion for f about a gives us that
X f (b) = f (a) − f (a)f 0 (a)f 0 (a) +
i=2deg
i
−f (a)f 0 (a) f (i) (a)/i!
f
As i! | f (i) (a) for every i and pij | (−f (a))i , we have
X i=2deg
−f (a)f 0 (a)
i
f (i) (a)/i! ≡ 0(mod p2j )
f
which in turn implies
f (b) ≡ f (a) − f (a)f 0 (a)f 0 (a)(mod p2j )
Now, by choice of f 0 (a), f 0 (a)f 0 (a) = 1 + kpj for some integer k and therefore
f (a)f 0 (a)f 0 (a) = f (a)(1 + kpj ) ≡ f (a)(mod p2j )
as pj | f (a) ⇒ p2j | kf (a)pj . Therefore,
f (b) ≡ f (a) − f (a) ≡ 0(mod p2j )
1
Q 5 (2.6(10)). Let p be an odd prime, and suppose that a 6≡ 0(mod p). Show that if the congruence
x2 ≡ a(mod pj ) has a solution when j = 1, then it has a solution for all j.
Proof. Suppose b(mod p) is a solution. We claim the derivative of the polynomial x2 − a (= 2x) does not
vanish at b(mod p). This is because if b2 ≡ a(mod p), then p - b2 and therefore p - b. As p is odd, this implies
p - 2b. By Hensel’s lemma, this solution can be lifted to a solution modulo pj for every j.
Q 6 (2.7(1)). Reduce the following congruences to equivalent congruences of degree ≤ 6:
(a) x11 + x8 + 5 ≡ 0(mod 7)
(b) x20 + x13 + x7 + x ≡ 2(mod 7)
(c) x15 − x10 + 4x − 3 ≡ 0(mod 7)
Proof.
x11 + x8 + 5 = (x7 − x + x)x4 + (x7 − x + x)x + 5 = x5 + x2 + 5 + (x7 − x)(x4 + x) ≡ x5 + x2 + 5(mod 7)
x20 +x13 +x7 +x−2 = (x14 −x2 +x2 )x6 +(x7 −x+x)x6 +(x7 −x+x)+x−2 = x8 +x7 +2x−2(mod 7) ≡ x2 +3x−2(mod 7)
x15 − x10 + 4x − 3 = (x14 − x2 + x2 )x − (x7 − x + x)x3 + 4x − 3 ≡ −x4 + x3 + 4x − 3(mod 7)
Q 7 (2.7(3)). Prove that the congruence x14 + 12x2 ≡ 0(mod 13) has 13 solutions and so it is an identical
congruence.
Proof.
x14 + 12x2 = x14 − x2 + 13x2 = x(x13 − x) + 13x2 ≡ 0(mod 13)
Q 8 (2.7(10)). Write 1/1 + 1/2 + · · · + 1/(p − 1) = a/b with (a, b) = 1. Show that p2 | a if p ≥ 5.
σ
p−2
Proof. This sum equals (p−1)!
. This tells us that bσp−2 = a(p − 1)!. Now we are given that p ≥ 5, so
2
2
p | σp−2 , and therefore p | a(p − 1)!. But (p2 , (p − 1)!) = 1, so this tells us that p2 | a.
Q 9 (2.7(11)). Let p be a prime, p ≥ 5, and suppose that the numbers σj are as in (2.7). Show that if
σp−2 ≡ pσp−3 (mod p3 ).
Proof. We have
pp−2 − σ1 pp−3 + · · · − σp−4 p2 + σp−3 p − σp−2 = 0
As p | σi for 1 ≤ i ≤ p − 2, we have p3 | σp−2−i pi for i ≥ 2. p ≥ 5 gives us p3 | pp−2 . Summing these with
the appropriate signs gives us
pp−2 − σ1 pp−3 + · · · − σp−4 p2 ≡ 0(mod p3 )
These two together give us σp−2 ≡ pσp−3 (mod p3 ).
Q 10 (2.7(12)). Show that if p ≥ 5 and m is a positive integer then
mp−1
p−1
≡ 1(mod p3 ).
Proof. Let f (x) = (x − 1)(x − 2) · · · (x − p + 1).
f (mp) = (mp−1)(mp−2) · · · (mp−p+1) = (mp)p−1 −σ1 (mp)p−2 +σ2 (mp)p−3 −· · ·−σp−2 (mp)+σp−1 ≡ σp−1 (mod p3 )
We have this congruence because p2 | σp−2 and therefore p3 | −σp−2 (mp), and p3 | σi (mp)p−i−1 for 1 ≤ i ≤
p − 3 (as p | σi for i < p − 1) and p3 | (mp)p−1 . This tells us that
mp − 1
(p − 1)!
= f (mp) ≡ (p − 1)!(mod p3 )
p−1
As ((p − 1)!, p3 ) = 1, we have
mp − 1
≡ 1(mod p3 )
p−1
2