MATH 115B SOLUTION SET VI
JUNE 11, 2007
(1) Show that there are infinitely many even integers
n
with the property that both
n
+ 1
and (
n/
2) + 1 are perfect squares. Exhibit two such integers.
Solution:
For an integer
n
, suppose that
n
+ 1 =
u
2
and (
n/
2) + 1 =
v
2
for some
u
and
v
. Then
u
2

2
v
2
= (
n
+ 1)

(
n
+ 2) =

1
.
Now
√
2 = [1;
2], and the convergents
C
2
k
=
p
2
k
/q
2
k
orf
√
2 provide solutions
u
=
p
2
k
,
v
=
q
2
k
of the equation
u
2

2
v
2
=

1. Thus there are infinitely many integers
n
for which
n
+ 1 and (
n/
2) + 1 are boths quares. Two such
n
are 48 and 1680.
(2) Find the fundamental solutions of the following equations:
(i)
x
2

29
y
2
= 1;
(ii)
x
2

41
y
2
= 1.
Solution:
(i) Observe that
√
29 = [5;
2
,
1
,
1
,
2
,
10] has period 5. Hence a fundamental solution of the
given equation is obtained from the convergent
C
9
of
√
29, namely
C
9
= 9801
/
1820 gives
x
= 9801,
y
= 1820.
(ii) We have that
√
41 = [6;
2
,
2
,
12] has period 3. The convergent
C
5
= 2049
/
320 gives
the fundamental solution
x
= 2049,
y
= 320 of the given equation.
(3)(a) Prove that whenever the equation
x
2

dy
2
=
c
is soluble, then it has infinitely
many solutions.
[Hint: If
u
,
v
satisfy
x
2

dy
2
=
c
and
r
,
s
satisfy
x
2

dy
2
= 1, then
(
ur
±
dvs
)
2

d
(
us
±
vr
)
2
= (
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 Fall '09
 Integers, Solubility, Prime number, equation x2

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