3. Compute
φ
(
n
) = (
p

1)(
q

1), the
totient
of
n
. This is the number of numbers among
1
,
2
,...,n

1 which have no factor in common with
n
.
4. Choose an exponent
e
such that 1
< e < φ
(
n
), where
e
and
φ
(
n
) are relatively prime.
e
is the
encoding exponent
.
5. Find
d
such that
de
≡
1
(mod
φ
(
n
)), by either trial and error or the Euclidean
algorithm.
d
is called the
decoding exponent
.
Then the
public key
(which is used to encode messages to be sent to you) is the pair of
numbers (
n,e
); the
private key
is the single number
d
. To encrypt the message
W
, compute
C
=
W
e
(mod
n
), and send that. To decode the ciphertext
C
, compute
C
d
(mod
n
).
This will be
W
.
Question:
Why do we choose a decoding exponent
d
such that
de
≡
1
(mod
φ
(
n
))? In
particular, what theorem must we use to prove that this is the correct decoding exponent,
and why?
Solution.
The reason for this choice of decoding exponent is that decoding should re
verse encoding: that is, after encoding and then decoding we should get back the original
message. Encoding and then decoding corresponds to raising the original message
W
to the
de
power and reducing
(mod
n
). Euler’s theorem allows us to ignore multiples of
φ
(
n
) in
the exponent;
de
is one more than a multiple of
φ
(
n
), and so
W
de
≡
W
(mod
n
).
10.
We saw two proofs that
√
2 is irrational. Explain why
one of these proofs
fails to
3
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View Full Documentshow that
√
4 is irrational. (The answer should be speciﬁc to one of the proofs we saw in
class; the observation that
√
4 is rational, by itself, does not suﬃce.)
Solution.
In one proof that we gave, we wrote
√
2 =
c/d
, which we rearranged to get
c
2
= 2
d
2
. We then observed that
c
2
is divisible by 2 and so
c
must be divisible by 2, which
is why the proof worked. But if
c
2
is divisible by 4, then
c
is not necessarily divisible by 4.
In the other proof we gave, we considered the prime factorizations of
c
and
d
in
c
2
= 2
d
2
.
On the lefthand side, there must be an even number of twos in the prime factorization; on
the righthand side there must be an odd number of twos. But this does not work when 2 is
replaced with 4 because now both sides have an even number of twos.
END OF EXAM
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 Summer '09
 Lugo
 Math, Solution., Bank identification number, prime factorizations

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