today’s writing project.
i played guitar (and even talked a little)
in church today, too, so it’s been
a pretty productive day for a sunday.
(i suppose. now that i think about it,
monday morning deadlines have led me
to many a *highly* productive sunday
is the actual *work*…)
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Let f(x) = x^3 + x + 57.
(Find all x s.t.
f(x)\equiv 0 (mod 125).
)
Since 125=5^3, we begin by working “mod 5”:
f(0) = 57 == 2
f(1) = 59 == 4
f(2) = 67 == 2
f(3) = 87 == 4
f(4) == f(-1) == 50 == 0 (mod 5).

So f has exactly one “mod-5 root”
(namely 4) and we consider
f'(4) = 3(4)^2 + 5 = 21 == 1 \not == 0 (mod 5).
This means that 4 is a *non-singular* root.

Hensel’s Lemma (HL)
now tells us that there is
exactly one root of f (mod 125).
But now, since
f(4) = 4^3 + 4 + 57 = 125,
we are done:
x=4
is the *only* solution to f(x)==0 (mod 125).
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Let g(x) = x^3 + 10x^2 + x + 3.
(Find all x s.t.
g(x)\equiv 0 (mod 27).
)
Since 27=3^3, we begin by working “mod 3”:
g(0)==0
g(1)==0
g(2)==2 (mod 3).
Next, we’ll compute g'(x) = 3x^2 + 20x + 1
and evaluate it at each of our “mod 3” roots:
g'(0) =1 and g'(1) = 24 == 0 (mod 3).

The root at 1 is *singular*
If “1” were lifted to a mod-9 root, x,
we would have x \in {1 +k*3} = {1, 4, 7}.
But
g(1) = 15 == 6,
g(4) = 231 == 1, and
g(7) = 843 == 6 (mod 9),
so there is *no* mod-9 root of f
satisfying x == 1 (mod 3)
(and so certainly no such mod-27 root).

The root at 0 is *non*-singular.
Since g'(0) = 1, its “mod-3 inverse” is
\bar{g'(0)} = 1.
“Plugging in” on “Hensel’s Formula”
(we have a_j = a_1 = a = 0;
our “f” is called “g”)
a_{j+1} = a_j – f(a_j)\bar{f'(a)}
(Display 2.6 of [NZM]) gives us
a_2 = 0 – 3(1) == -3 == 6 (mod 9).

Repeating the process,
a_3 = a_2 – g(a_2)\bar{g'(a)}
a_3 = 6 – 585(1)
a_3 = -579
a_3 = -(21*27 + 12).
a_3 == -12 == 15 (mod 27).
Our one-and-only solution
to f(x)==0 (mod 27) is x = 15.

f(15) = 5643 = 209*27 == 0 (mod 27).)
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