Midterm. Mathematics 312, Section 101
Thursday, October 21, 2010. Instructor: Z. Reichstein.
Problem 1:
(6 marks) Use mathematical induction to show that 2
n
> n
2
for every
integer
n
≥
5.
Solution:
Base case. For
n
= 5, 2
n
= 32 and
n
2
= 25, so the desired inequality
holds.
For the induction step assume
(1)
2
n
> n
2
for some
n
≥
5. We need to show that 2
n
+1
>
(
n
+ 1)
2
. Multiplying both sides of (1)
by 2, we obtain 2
n
+1
>
2
n
2
. It thus remains to show that
2
n
2
≥
(
n
+ 1)
2
for every
n
≥
5. Indeed,
2
n
2
>
36
25
n
2
= (1 +
1
5
)
2
n
2
≥
(1 +
1
n
)
2
n
2
= (
n
+ 1)
2
,
as claimed. This completes the proof.
Problem 2:
(6 marks) Find
all
integer solutions of the linear equation
ax
+ (3
a
+ 1)
y
= 1
.
Here
a
is a given positive integer and
x
,
y
are unknowns.
Solution:
One solution is clearly
x
0
=

3,
y
0
= 1. Since it exists, gcd(
a,
3
a
+ 1)
divides 1 and hence, equals 1.
The general solution is thus
x
=

3 + (3
a
+ 1)
n
,
y
= 1

an
, as
n
ranges over the integers.
Remark: The fact that gcd(3
a
+ 1
, a
) = 1 can also be checked directly as follows:
gcd(3
a
+ 1
, a
) = gcd(3
a
+ 1

3
a, a
) = gcd(1
, a
) = 1. The above slution does not rely
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 Winter '10
 ZinovyReichstein
 Math, Remainder, Mathematical Induction, Natural number, Greatest common divisor, Euclidean algorithm

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