Download MT 430 Intro to Number Theory PROBLEM SET 3 Due Thursday 2

MT 430 Intro to Number Theory
PROBLEM SET 3
Due Thursday 2/14
Remark. Here, as in class, whenever working modulo m, we use notation x to
denote a multiplicative inverse of x modulo m. Whenever (x, m) = 1, x is the
unique residue modulo m such that xx ≡ 1 (mod m).
Problem 1. Let p be an odd prime number. Show that the congruence
x2 ≡ 1
(mod pn )
has only the two solutions x ≡ 1 (mod pn ) and x ≡ −1 (mod pn ).
Solution: The statement x2 ≡ 1 (mod pn ) is equivalent to
pn | (x2 − 1) = (x − 1)(x + 1).
Since (x + 1) − (x − 1) = 2, the numbers (x + 1) and (x − 1) cannot be divisible
by p simultaneously. If (x − 1) is not divisible by p, then since p is a prime and pn
divides the product (x − 1)(x + 1), we must have pn | x + 1. Similarly, if (x + 1) is
not divisible by p, we must have pn | x − 1.
Problem 2. Find all x such that


x ≡ 1 (mod 4)
x ≡ 2 (mod 3)


x ≡ 5 (mod 7)
We apply the algorithmic version of Chinese Remainder Theorem from class. Take
b1 = 21 = 1 = 1 (mod 4), b2 = 28 = 1 = 1 (mod 3), and b3 = 12 = 5 = 3 (mod 7).
Then
x = 1 × b1 × 21 + 2 × b2 × 28 + 5 × b3 × 12 = 21 + 56 + 180 ≡ 5
(mod 84)
is the unique solution of the system modulo 84 = 4 × 3 × 7.
Problem 3. Solve the congruence x3 + 2x − 3 ≡ 0 (mod 45).
Solution: Note that 45 = 5 × 32 . First, we solve separately the congruence x3 +
2x − 3 ≡ 0 modulo 5 and modulo 32 . The solutions of x3 + 2x − 3 ≡ 0 (mod 5) are
x ≡ 1, 3 (mod 5).
To find solutions modulo 32 , we first observe that modulo 3, x3 +2x−3 ≡ x+2x−
3 ≡ 3x − 3 ≡ 0 (mod 3) by Fermat’s little theorem. Hence all x ≡ 0, 1, 2 (mod 3)
are all solutions modulo 3. We then compute that (x3 + 2x − 3)0 = 3x2 + 2 ≡ 2
(mod 3). Hence by Hensel’s lemma, every solution modulo 3 lifts to a unique
solution modulo 9. Since 2 = 2 (mod 3), we see that
• x ≡ 0 (mod 3) lifts to x ≡ 0 − 2f (0) = 0 − 2(−3) = 6 (mod 9)
• x ≡ 1 (mod 3) lifts to x ≡ 1 − 2f (1) = 1 − 0 = 1 (mod 9)
• x ≡ 2 (mod 3) lifts to x ≡ 2 − 2f (2) = 2 − 18 = 2 (mod 9)
We summarize that solutions of x3 + 2x − 3 ≡ 0 (mod 9) are x ≡ 1, 2, 6 (mod 9).
Combining these with the solution x ≡ 1, 3 (mod 5) of x3 + 2x − 3 ≡ 0 (mod 5)
and applying Chinese Remainder Theorem, we obtain 6 solutions modulo 45:
x ≡ 1, 28, 11, 38, 6, 33
1
(mod 45).
2
PROBLEM SET 3
Problem 4. Solve x5 + x4 + 1 ≡ 0 (mod 34 ).
Hint: Hensel’s lemma is inapplicable here because f 0 (1) = 3 ≡ 0 (mod 3). You
can still use the unique solution modulo 3 to look for solutions modulo 9.
Solution: The only solution of x5 + x4 + 1 ≡ 0 (mod 3) is x = 1. Hence the only
candidates for solutions of x5 + x4 + 1 ≡ 0 (mod 32 ) are x = 1, 1 + 3 = 4, and
x = 1 + 6 = 7. We check that
15 + 14 + 1 ≡ 3
5
(mod 9)
4
4 + 4 + 1 ≡ 3 (mod 9)
75 + 74 + 1 ≡ 3 (mod 9)
5
4
Therefore, the x + x + 1 ≡ 0 has no solutions modulo 9 and thus no solutions
modulo 34 .
Problem 5 (Edited on 2/12). Let p be an odd prime and suppose that a is an
integer such that x2 ≡ a (mod p) has a solution and a 6≡ 0 (mod p). Show that
x2 ≡ a (mod pn ) has a solution for all integers n ≥ 1.
Solution: The proof is by induction. The base case of n = 1 is given to us. Suppose
b is a solution of x2 − a ≡ 0 (mod pn ). Since (x2 − a)0 = 2x and 2b 6≡ 0 (mod p),
Hensel’s lemma says that b lifts to a solution of x2 − a ≡ 0 (mod pn+1 ).
Remark. Euler’s criterion says that for a 6≡ 0 (mod p) the equation x2 ≡ a
(mod p) has a solution if and only if
x
p−1
2
≡1
(mod p).
Assuming Euler’s criterion, the old statement of Problem 5 is equivalent to the new
statement.