Math 4430
Problem set #7
Due Monday, March 20
Exercise 1. Find a solution to x3 + 2x + 2 ≡ 0 (mod 125).
Solution. Set f (x) = x3 + 2x + 2. We start by noting that α1 = 3 satisfies
f (α1 ) ≡ 0
f 0 (α1 ) ≡ −1
(mod 5),
(mod 5).
In particular t = −1 satisfies t · f 0 (α1 ) ≡ 1 (mod 5). Applying Hensel’s lemma, we compute
α2 = α1 − t · f (α1 ) ≡ 13
(mod 52 )
α3 = α2 − t · f (α2 ) ≡ 113
(mod 53 ).
Thus x = 113 is a solution to x3 + 2x + 2 ≡ 0 (mod 125).
Exercise 2. Find a solution to x3 ≡ 7 (mod 200).
Solution. Set f (x) = x3 − 7, and note that 200 = 8 · 25. Obviously
f (−1) ≡ 0
(mod 8).
By trial and error we find that f (3) ≡ 0 (mod 5), and we can use Hensel’s lemma to then
find
f (18) ≡ 0 (mod 25).
Now we can use the Chinese remainder theorem to solve the simultaneous congruences
x ≡ −1
(mod 8)
x ≡ 18
(mod 25).
Indeed, the usual method produces the solution x ≡ 143 (mod 200), which therefore solves
f (x) ≡ 0 (mod 200). In other words,
1433 ≡ 7
(mod 200).
Exercise 3. Find a solution to x2 ≡ 17 (mod 27 ).
Solution. Here Hensel’s lemma does not apply, but we have a similar method from class
for finding a solution. Start with α3 = 1, so that α32 ≡ 17 (mod 23 ). Now define
αk+1 ≡ αk +
17 − αk2
2
(mod 2k+1 ),
so that
17 − 1
≡ 9 (mod 24 )
2
17 − 92
α5 = 9 +
≡ 9 (mod 25 )
2
17 − 92
≡ 41 (mod 26 )
α6 = 9 +
2
17 − 412
≡ 105 (mod 27 ).
α7 = 41 +
2
α4 = 1 +
This gives the desired solution 1052 ≡ 17 (mod 27 ).
Exercise 4. Express each of the following rational numbers as a continued fraction:
(a) 56/9,
(b) 103/19,
(c) −3/21,
(d) −13/11.
Solution. The continued fraction expansions are
(a) 56/9 = CF(6; 4, 2)
(b) 103/19 = CF(5; 2, 2, 1, 2)
(c) −3/21 = CF(−1; 1, 6)
(d) −13/11 = CF(−2; 1, 4, 2)
Exercise 5. Define a sequence x1 , x2 , x3 , . . . by
xk = CF(2; 4, . . . , 4).
| {z }
k times
Assuming that the limit x = limk→∞ xk exists, what it is?
Solution. The sequence yk = 2 + xk = CF(4; 4, . . . , 4) converges to y = 2 + x and satisfies
| {z }
k times
yk+1 = 4 +
1
.
yk
Passing to the limit in this expression gives
1
y =4+ ,
y
which we can rewrite as y 2 − 4y − 1 = 0. Using the quadratic formula we find
√
√
4 + 20
= 2 + 5.
y=
2
√
Hence x = y − 2 = 5.
Alternatively, you could use the relation
xk+1 = 2 +
take the limit to get x = 2 +
1
2+x ,
1
,
2 + xk
rewrite this as x2 = 5, and deduce that x =
√
5.
Exercise 6. Define a sequence x1 , x2 , x3 , . . . by
x1 = CF(1; 1)
x2 = CF(1; 1, 2)
x3 = CF(1; 1, 2, 1)
x4 = CF(1; 1, 2, 1, 2)
x5 = CF(1; 1, 2, 1, 2, 1)
···
Assuming that the limit x = limk→∞ xk exists, what it is?
Solution. If we only consider the subsequence
yk = x2k = CF(1; 1, 2, . . . , 1, 2)
|
{z
}
k times
we must get the same limit, x. These satisfy
yk+1 = 1 +
1
1+
1
1+yk
,
and taking the limit gives
x=1+
1
1+x
3 + 2x
=1+
=
.
1
2+x
2+x
1 + 1+x
This implies x(2 + x) = 3 + 2x, which we rewrite as x2 = 3. Thus x =
√
3.
Exercise 7. Suppose f (x) is a polynomial with integer coefficients, p is a prime, and α ∈ Z
satisfies
f (α) ≡ 0 (mod p), f 0 (α) 6≡ 0 (mod p).
(a) Prove that there is an αk ∈ Z such that f (αk ) ≡ 0 (mod pk ) and αk ≡ α (mod p).
(b) Prove that αk is uniquely determined (mod pk ) by these properties.
Hint: for part (a) use Hensel’s lemma. For part (b), use the Taylor expansion
f (αk + x) = f (αk ) + xf 0 (αk ) + (terms involving x2 , x3 , . . .).
Solution. (a) As in Hensel’s lemma, set α0 = α, choose a t ∈ Z such that t · f 0 (α0 ) ≡ 1
(mod p), and define αk recursively by
αk ≡ αk−1 − t · f (αk−1 )
(mod pk ).
Hensel’s lemma tells us that f (αk ) ≡ 0 (mod pk ), and so the only thing to check is that
αk ≡ α (mod p). We do this by induction. The base case k = 0 is trivial, since α0 = α.
Now assume that αk−1 ≡ α (mod p), so that
αk ≡ αk−1 − t · f (αk−1 )
(mod pk ) =⇒ αk ≡ αk−1 − t · f (αk−1 )
=⇒ αk ≡ αk−1
(mod pk−1 )
=⇒ αk ≡ αk−1
(mod p)
=⇒ αk ≡ α
(mod pk−1 )
(mod p).
(b) Suppose that βk ∈ Z/pk Z also satisfies f (βk ) ≡ 0 (mod pk ) and βk ≡ α (mod p). If
we set x = βk − αk then
x ≡ α − α ≡ 0 (mod p),
so that p | x. Using the Taylor expansion of the hint, we compute
0 ≡ f (βk )
(mod pk )
≡ f (αk + x)
(mod pk )
≡ f (αk ) + xf 0 (αk ) + terms involving x2 , x3 , . . . (mod pk )
≡ x · f 0 (αk ) + terms involving x, x2 , . . .
(mod pk ).
On the other hand, we already know that
f 0 (αk ) + terms involving x, x2 , . . . ≡ f 0 (αk )
0
≡ f (α)
6≡ 0
(mod p)
(mod p)
(mod p),
and so f 0 (αk ) + terms involving x, x2 , . . . is relatively prime to p. This implies that
f 0 (αk ) + terms involving x, x2 , . . . ∈ (Z/pk Z)× .
Returning to
0 ≡ x · f 0 (αk ) + terms involving x, x2 , . . .
we deduce that x ≡ 0 (mod pk ), and hence αk ≡ βk (mod pk ).
(mod pk ),