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The problem is therefore to deal with polynomials F ( x ) having a small solution but
whose coefficients are not small. Coppersmith's idea (in the formulation of Howgrave-
Graham [ 268 ]) is to build from F ( x ) a polynomial G ( x ) that still has the same solution x 0 ,
but which has coefficients small enough that the above logic does apply.
=
·
=
Example 19.1.1 Let M
17
19
323 and let
x 2
F ( x )
=
+
33 x
+
215 .
We want to find the small solution to F ( x )
0(mod M ) (in this case, x 0 =
3 is a solution,
but note that F (3)
).
We seek a related polynomial with small coefficients. For this example
=
0 over
Z
9 x 2
=
+
=
G ( x )
9 F ( x )
M ( x
6)
26 x
3
satisfies G (3)
=
0. This root can be found using Newton's method over
R
(or even the
quadratic formula).
We introduce some notation for the rest of this section. Let M,X
∈ N
and let F ( x )
=
i = 0 a i x i
∈ Z
[ x ]. Suppose x 0 ∈ Z
is a solution to F ( x )
0(mod M ) such that
|
x 0 |
<X .
We associate with the polynomial F ( x ) the row vector
( a 0 ,a 1 X,a 2 X 2 ,
,a d X d ) .
b F =
···
(19.1)
Vice versa, any such row vector corresponds to a polynomial. Throughout this section we
will interpret polynomials as row vectors, and row vectors as polynomials, in this way.
Theorem 19.1.2 (Howgrave-Graham [ 268 ]) Let F ( x ) ,X,M,b F be as above (i.e., there
is some x 0 such that
<M/ d
|
x 0 |≤
X and F ( x 0 )
0(mod M ) ). If
b F
+
1 then
F ( x 0 )
0 .
Proof Recall the Cauchy-Schwarz inequality ( i = 1 x i y i ) 2
=
( i = 1 x i )( i = 1 y i )for
x i ,y i ∈ R
. Taking x i
0 and y i =
1for1
i
n one has
n
n
n
x i .
x i
i = 1
i = 1
Now
d
d
d
i = 0 |
i = 0 |
a i x i 0
i
X i
|
F ( x 0 )
|=
a i ||
x 0 |
a i |
i = 0
d
< d
1 M/ d
+
1
b F
+
+
1
=
M
where the third inequality is Cauchy-Schwarz, so
M<F ( x 0 ) <M .But F ( x 0 )
0(mod M ) and so F ( x 0 )
=
0.
= i = 0 a i x i be a monic polynomial. We assume that F ( x ) has at least one
solution x 0 modulo M such that
Let F ( x )
|
x 0 |
<X for some specified integer X .If F ( x )isnot
 
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