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Math 111 Midterm Solutions
Prof. Yehuda Shalom
TA: Darren Creutz
Date: 4 May 2009
Problem 1.
Compute
d
=
gcd
(29 + 4
i,
1 + 2
i
) in
Z
[
i
], the Gaussian integers (
i
=
√

1). Find Gaussian integers
n
and
m
such that (29 + 4
i
)
n
+ (1 + 2
i
)
m
=
d
.
Solution.
First observe that
29 + 4
i
1 + 2
i
=
(29 + 4
i
)(1

2
i
)
(1 + 2
i
)(1

2
i
)
=
29 + 8

58
i
+ 4
i
5
= 7

10
i
+
2

4
i
5
= 7

10
i
+
2
1 + 2
i
Therefore
gcd
(29 + 4
i,
1 + 2
i
) =
gcd
(1 + 2
i,
2) =
gcd
(2
,
1 + 2
i

2
i
) =
gcd
(2
,
1) = 1
Now tracing what we did, set:
29 + 4
i
= (1 + 2
i
)(7

10
i
) + 2
so
(29 + 4
i
)

(7

10
i
)(1 + 2
i
) = 2
and
1 + 2
i
=
i
*
2 + 1 =
i
*
[(29 + 4
i
)

(7

10
i
)(1 + 2
i
)] + 1
so

i
(29 + 4
i
) + [
i
(7

10
i
) + 1](1 + 2
i
) = 1
and so
n
=

i
and
m
= 11 + 7
i
will work:

i
(29 + 4
i
) + (11 + 7
i
)(1 + 2
i
) =

29
i
+ 4 + 11 + 22
i
+ 7
i

14 = 1
Problem 2.
Find all integer solutions to the following system of congruences:
x
= 2 mod 7
3
x
= 1 mod 5
x
= 7 mod 10
Solution.
First since 3

1
mod 5 = 2 replace the second equation by
x
= 2 mod 5. Since 7 is relatively prime to
5 and 10 and since
gcd
(5
,
10) = 5 divides 7

2 = 5, the Chinese Remainder Theorem applies. There is then a
unique solution
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