1
(a)
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
e x t r a o d i n y b c f g h j k l m p q s u v w z(b)
Twxal matqlnpw
(c)
Coec hv aevj
2
Eulfsraxetamr gjyrtwb
3
i. Frequency table
| A | F | G | H | J | K | M | N | O | P | S | U | V | X | Y | Z |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 2 | 5 | 2 | 4 | 4 | 1 | 4 | 1 | 3 | 1 | 1 | 4 | 4 | 2 | 4 |
3.1 ii.
By frequency analysis the remaining substitutions needed is
| G | K | N | J | P | Z | F | H | U | O | A | M |
|---|---|---|---|---|---|---|---|---|---|---|---|
| i | t | e | o | h | s | a | c | y | g | f | u |
to get
neitherofthoseisitsmaincryptographicpurpose4
(a)
\(1519 = 49\cdot 31 + 0\), \((q,r) = (49,0)\).
\(11105 = 64\cdot 171 + 161\), \((q,r) = (64,161)\).
\(-8747891 = -3770\cdot 2321 + 2279\), \((q,r) = (-3770,2279)\).
(b)
trivial, \(31 = \gcd(1519, 31) = 31 + 0\)
\(\gcd(11105, 171) = 1 = 17 \cdot 11105 - 1104 \cdot 171\)
\(\gcd(-8747891, 2321) = 1 = (-1050) \cdot (-8747891) - 3957469 \cdot 2321\)
5
Let \(k,l\) be integers such that \(ak = b\) and \(bl = c\), then \(akl = c\), so \[ bx + cy^2 = akx + akly^2 = a(kx + kly^2) \] which shows the divisibility.
6
Let \(x,y\) be integers such that \[ abx + c^2y = 1. \] \(\gcd(a,c) = 1\) because \(a(bx) + c(cy) = 1\), similarly \(\gcd(b,c) = 1\) because \(b(ax) + c(cy) = 1\).