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MCS 425 Midterm Exam Solutions — Spring 2008
1. [ 4 points] An affine cipher E ,
( x ) = x + (mod 26) encrypts plaintext er as
ciphertext J
, where
represents some ciphertext letter. Note the 26 letters correspond to
the integers {0,1,...,25} as follows:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
a b c d e f g h i j k l m n o p q r s t u v w x y z
a) [3 points] What is
, i.e., what is the encryption of r?
From E
,
( e ) = J and E
,
( r ) =
,
we obtain
4 + 9 (mod 26),
(mod 26).
Subtracting the first equation from the second gives
9 (mod 26).
Since gcd (,26) = 1, must be odd, i.e., 1 (mod 2). It follows that
13 13 (mod 26). Then
9 + 139 + 13 22. Thus r is encrypted
to W.
b) [ 1 points] Could the cipher described above encrypt b as C? Why or why not?
Yes, it could. To see this, we have to show that the simultaneous equations
4 + 9 (mod 26),
17 + 22 (mod 26),
+ 2 (mod 26),
have a solution.
The second equation follows from the first, so we can ignore it. Subtracting
the third equation from the first gives
3 7 (mod 26).
Thus 3
1
7 9 7 63 11 (mod 26). From the third equation,
2 2 11 9 17 (mod 26).
Thus there is a solution, 11 (mod 26) and 17 (mod 26).
2. [5 points] Use the fact that
2
2
(mod 36503)
to produce a nontrivial factorization of 36503. Show your work, and use only methods
applicable even with very large integers.
Note 903
/ 481 (mod 36503), so by a major theorem proven in class,
36503 = gcd ( 36503, 90 3 481 ) gcd ( 36503, 903+481)
provided that one (and hence both) of 481 and 903 are relatively prime to
gcd ( 36503, 90 3 481) = gcd ( 36503, 422) is computed as follows:
So gcd ( 36503, 422) = 2 1 1. 36503 / 21 1 = 173, so 36503 = 21 1 173.
3. [ 6 points] In this problem, show all your work, and use only techniques that can be
used even with very large integers.
a) [ 2 points] Show that one of 3 or 5 is a quadratic residue mod 23, and the other is a
nonresidue.
(2 3 1 ) / 2 = 11. a is a quadratic residue mod 23 if and only if a
11
1 (mod 23).
3
2
9 (mod 23)
3
2
2
9
2
81 12 (mod 23)
3
2
3
12
2
144 6 (mod 23)
3
11
3
2
3
3
2
3 6 9 3 162 1 (mod 23)
So 3 is a quadratic residue mod 23.
5
2
25 2 (mod 23)
5
2
2
2
2
4 (mod 23)
5
2
3
4
2
16 (mod 23)
5
11
5
2
3
5
2
5 16 2 5 160 1 (mod 23)
So 5 is a non-residue mod 23.
b) [ 2 points] Compute the square roots of 3 or 5 (the one that is a residue) mod 23.
Since 23 3 (mod 4), and since we know that 3 is a quadratic residue mod
23, the square roots of 3 mod 23 must be 3
(23+1)/ 4
6
. Using the values
of 3
2
and 3
2
2
above, we compute 3
6
2
2
2
12 9 108 16 (mod 23).
So the square roots of 3 mod 23 are 16. (They may also be written as 7 .)
c) [ 2 points] Using only your result in part (a), and without performing any more
computation, decide whether 15 is a quadratic residue mod 23.
We know that ( residue ) ( non-residue ) = ( non-residue ). Since 3 is a residue
mod 23 and 5 is a non-residue, 15 is a non-residue mod 23.
