128
Solutions to Exercises
3) Let
k
::::: 2 and assume
P(k)
is true. We must prove that
P(k
1)
is true. So we suppose that
aI,
... ,
akl
are
k
1 nonnegative numbers. The idea is now to apply
P( k)
to the
k
numbers
aI,
a2,
... , akl, k.yal··· akl
since the
kth
root of the product of these numbers is
Thus,
and so
k/
_1
k/
H_1
y
a
l···ak_l(al···ak_d
k

1
=
y(al
... akl)
k1
~(al
...
ak_I)6
=
(al··
.ak_d
l
/
k

l
=
k.yal·· ·akl·
k.yal··· akl
=
~(al·.·
akl) k .yal
...
akl
al
+
a2
+ .
.. +
akl
+
k.yal··· akl
::;
k
'
usingP(k)
(
1)
al+···+akl
k.yal
...
akl
1 
k::;
k
(
k1)
al+···+akl
k.yal
...
akl
k::; k
al
+ .
.. +
akl
k1
which is
P(k

1),
as desired.
41. (With thanks to M. Oleson) (a) Assume to the contrary that
mn
 m 
n
=
am
+
bn
where
a
and
b
are nonnegative integers. Then
m(n

1 
a)
=
n(l
+
b),
so
n
I
m(n

1 
a).
Since m and
n
are
relatively prime,
n
I
(n
 1 
a)
by Corollary 4.2.10, which is impossible since 0
<
n
 1 
a
<
n.
(b) There exist integers
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 Summer '10
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 Graph Theory, Addition, Negative Numbers, Prime number, 1 K, kl, cent stamps

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