(1) Prove that for any
n
≥
1
,
(1)
x
n
−
y
n
=(
x
−
y
)(
x
n
−
1
+
x
n
−
2
y
+
···
+
y
n
−
1
)
.
Solution
: The proof will be
by induction
on
n
. One convenient way to do it, which simpliFes
the algebra involved, is to Frst show that for any
n
≥
1,
(2)
1
−
t
n
=(1
−
t
)(1 +
t
+
···
+
t
n
−
1
)
.
Then setting
t
=
y/x
we get (1) after multiplication by
x
n
. So let’s prove (2). ±or
n
=1the
equality (2) is clearly true. Assume now (induction hypothesis) that it holds for
n
=
k
, so that
(3)
1
−
t
k
=(1
−
t
)(1 +
t
+
···
+
t
k
−
1
)
,
and let’s try to show that (2) also holds for
n
=
k
+ 1. And indeed,
1+
t
+
···
+
t
k
−
1
+
t
k
=(1+
t
+
···
+
t
k
−
1
)+
t
k
=
1
−
t
k
1
−
t
+
t
k
(by the induction hypothesis (3))
=
1
−
t
k
+(
t
k
−
t
k
+1
)
1
−
t
=
1
−
t
k
+1
1
−
t
,
as was to be shown
°
.
(2) Let
a
and
b
be nonzero integers such that
b
=
aq
+
r
, where
0
≤
r<a
. Prove that
(
b, a
)=(
a, r
)
. [As will be clear from the proof, the assumption on
r
is not needed here.]
Solution
: ±irst we show that
(4)
(
b, a
)d
iv
ides(
a, r
):
indeed, by deFnition (
b, a
) divides both
b
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 Fall '07
 Goren
 Algebra, Prime number, Divisor, Prime factor, Table of prime factors

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