A number theory problem by Rocco Dalto

Using $a \rightarrow 0, b \rightarrow 1, \ldots , z \rightarrow 25$ group the plaintext word and the enciphered text word into blocks of size 5. Using each plaintext block of size 5 form the plaintext matrix $P_{5 x 5}$ . Using each enciphered block of size 5 form the enciphered matrix $C_{5 x 5}$ , then $A_{5 x 5} * P_{5 X 5} \equiv C_{5 x 5} \bmod{26} \implies A_{5 x 5} \equiv C_{5 x 5} * P^{-1}_{5 x 5} \bmod{26}$ , where $P^{-1}_{5 x 5}$ is the inverse of $P_{5 x 5}$ .

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Let $P = \begin{bmatrix}{8} && {14} && {19} && {14} && {2} \\ {12} && {4} && {17} && {17} && {0} \\ {12} && {11} && {14} && {4} && {11} \\ {20} && {4} && {15} && {19} && {11} \\ {13} && {2} && {7} && {8} && {24}\end{bmatrix} \bmod{26}$

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Since $\det(P) = 5 \bmod{26}$ and $(5,26) = 1 \implies P^{-1} \bmod{26}$ exists.

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Set up the Augmented matrix $P|I$ :

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$P|I = \left[ \begin{array}{ccccc|ccccc} 8 & 14 & 19 & 14 & 2 & 1 & 0 & 0 & 0 & 0\\ 12 & 4 & 17 & 17 & 0 & 0 & 1 & 0 & 0 & 0\\ 12 & 11 & 14 & 4 & 11 & 0 & 0 & 1 & 0 & 0\\ 20 & 4 & 15 & 19 & 11 & 0 & 0 & 0 & 1 & 0\\ 13 & 2 & 7 & 8 & 24 & 0 & 0 & 0 & 0 & 1\\ \ \end{array} \right] \bmod{26}$

$Row_{2} \leftrightarrow Row_{3} \rightarrow$

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$\left[ \begin{array}{ccccc|ccccc} 8 & 14 & 19 & 14 & 2 & 1 & 0 & 0 & 0 & 0\\ 12 & 11 & 14 & 4 & 11 & 0 & 0 & 1 & 0 & 0\\ 12 & 4 & 17 & 17 & 0 & 0 & 1 & 0 & 0 & 0\\ 20 & 4 & 15 & 19 & 11 & 0 & 0 & 0 & 1 & 0\\ 13 & 2 & 7 & 8 & 24 & 0 & 0 & 0 & 0 & 1\\ \ \end{array} \right] \bmod{26}$

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$19 * Row_{2} \rightarrow$

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$\left[ \begin{array}{ccccc|ccccc} 8 & 14 & 19 & 14 & 2 & 1 & 0 & 0 & 0 & 0\\ 20 & 1 & 6 & 24 & 1 & 0 & 0 & 19 & 0 & 0\\ 12 & 4 & 17 & 17 & 0 & 0 & 1 & 0 & 0 & 0\\ 20 & 4 & 15 & 19 & 11 & 0 & 0 & 0 & 1 & 0\\ 13 & 2 & 7 & 8 & 24 & 0 & 0 & 0 & 0 & 1\\ \ \end{array} \right] \bmod{26}$

$$

$12 * Row_{2} + Row_{1}$

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$22 * Row_{2} + Row_{3}$

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$22 * Row_{2} + Row_{4}$

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$24 * Row_{2} + Row_{5}$

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$\left[ \begin{array}{ccccc|ccccc} 14 & 0 & 13 & 16 & 14 & 1 & 0 & 20 & 0 & 0\\ 20 & 1 & 6 & 24 & 1 & 0 & 0 & 19 & 0 & 0\\ 10 & 0 & 19 & 25 & 22 & 0 & 1 & 2 & 0 & 0\\ 18 & 0 & 17 & 1 & 7 & 0 & 0 & 2 & 1 & 0\\ 25 & 0 & 21 & 12 & 22 & 0 & 0 & 14 & 0 & 1\\ \ \end{array} \right] \bmod{26}$

$$

$11 * Row_{3} \rightarrow$

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$\left[ \begin{array}{ccccc|ccccc} 14 & 0 & 13 & 16 & 14 & 1 & 0 & 20 & 0 & 0\\ 20 & 1 & 6 & 24 & 1 & 0 & 0 & 19 & 0 & 0\\ 6 & 0 & 1 & 15 & 8 & 0 & 11 & 22 & 0 & 0\\ 18 & 0 & 17 & 1 & 7 & 0 & 0 & 2 & 1 & 0\\ 25 & 0 & 21 & 12 & 22 & 0 & 0 & 14 & 0 & 1\\ \ \end{array} \right] \bmod{26}$

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$13 * Row_{3} + Row_{1}$

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$20 * Row_ {3} + Row_{2}$

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$9 * Row_{3} + Row_{4}$

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$5 * Row_{3} + Row_{5}$

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$\left[ \begin{array}{ccccc|ccccc} 14 & 0 & 0 & 3 & 14 & 1 & 13 & 20 & 0 & 0\\ 10 & 1 & 0 & 12 & 5 & 0 & 12 & 17 & 0 & 0\\ 6 & 0 & 1 & 15 & 8 & 0 & 11 & 22 & 0 & 0\\ 20 & 0 & 0 & 6 & 1 & 0 & 21 & 18 & 1 & 0\\ 3 & 0 & 0 & 9 & 10 & 0 & 3 & 20 & 0 & 1\\ \ \end{array} \right] \bmod{26}$

$$

$Row_{4} \leftrightarrow Row_{5} \rightarrow$

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$\left[ \begin{array}{ccccc|ccccc} 14 & 0 & 0 & 3 & 14 & 1 & 13 & 20 & 0 & 0\\ 10 & 1 & 0 & 12 & 5 & 0 & 12 & 17 & 0 & 0\\ 6 & 0 & 1 & 15 & 8 & 0 & 11 & 22 & 0 & 0\\ 3 & 0 & 0 & 9 & 10 & 0 & 3 & 20 & 0 & 1\\ 20 & 0 & 0 & 6 & 1 & 0 & 21 & 18 & 1 & 0\\ \ \end{array} \right] \bmod{26}$

$$

$3 * Row_{4} \rightarrow$

$\left[ \begin{array}{ccccc|ccccc} 14 & 0 & 0 & 3 & 14 & 1 & 13 & 20 & 0 & 0\\ 10 & 1 & 0 & 12 & 5 & 0 & 12 & 17 & 0 & 0\\ 6 & 0 & 1 & 15 & 8 & 0 & 11 & 22 & 0 & 0\\ 9 & 0 & 0 & 1 & 4 & 0 & 9 & 8 & 0 & 3\\ 20 & 0 & 0 & 6 & 1 & 0 & 21 & 18 & 1 & 0\\ \ \end{array} \right] \bmod{26}$

$$

$23 * Row_{4} + Row_{1}$

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$14 * Row_{ 4} + Row_{2}$

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$11 * Row_{4} + Row_{3}$

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$20 * Row _{4} + Row_{5}$

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$\left[ \begin{array}{ccccc|ccccc} 13 & 0 & 0 & 0 & 2 & 1 & 12 & 22 & 0 & 17\\ 6 & 1 & 0 & 0 & 9 & 0 & 8 & 25 & 0 & 16\\ 1 & 0 & 1 & 0 & 0 & 0 & 6 & 6 & 0 & 7\\ 9 & 0 & 0 & 1 & 4 & 0 & 9 & 8 & 0 & 3\\ 18 & 0 & 0 & 0 & 3 & 0 & 19 & 22 & 1 & 8\\ \ \end{array} \right] \bmod{26}$

$$

$9 * Row_{5} \rightarrow$

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$\left[ \begin{array}{ccccc|ccccc} 13 & 0 & 0 & 0 & 2 & 1 & 12 & 22 & 0 & 17\\ 6 & 1 & 0 & 0 & 9 & 0 & 8 & 25 & 0 & 16\\ 1 & 0 & 1 & 0 & 0 & 0 & 6 & 6 & 0 & 7\\ 9 & 0 & 0 & 1 & 4 & 0 & 9 & 8 & 0 & 3\\ 6 & 0 & 0 & 0 & 1 & 0 & 15 & 16 & 9 & 20\\ \ \end{array} \right] \bmod{26}$

$$

$24 * Row_{ 5} + Row_{1}$

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$17 * Row_{5} + Row_{2}$

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$26 * Row_{5} + Row_{3}$

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$22 * Row_{5} + Row{4}$

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$\left[ \begin{array}{ccccc|ccccc} 1 & 0 & 0 & 0 & 0 & 1 & 8 & 16 & 8 & 3\\ 4 & 1 & 0 & 0 & 0 & 0 & 3 & 11 & 23 & 18\\ 1 & 0 & 1 & 0 & 0 & 0 & 6 & 6 & 0 & 7\\ 11 & 0 & 0 & 1 & 0 & 0 & 1 & 22 & 16 & 1\\ 6 & 0 & 0 & 0 & 1 & 0 & 15 & 16 & 9 & 20\\ \ \end{array} \right] \bmod{26}$

$22 * Row_{1} + Row_{2}$

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$25 * Row_{1} + Row_{3}$

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$15 * Row{1} + Row_{4}$

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$20 * Row_{1} + Row_{5}$

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$\left[ \begin{array}{ccccc|ccccc} 1 & 0 & 0 & 0 & 0 & 1 & 8 & 16 & 8 & 3\\ 0 & 1 & 0 & 0 & 0 & 22 & 23 & 25 & 17 & 6\\ 0 & 0 & 1 & 0 & 0 & 25 & 24 & 16 & 18 & 4\\ 0 & 0 & 0 & 1 & 0 & 15 & 17 & 2 & 6 & 20\\ 0 & 0 & 0 & 0 & 1 & 20 & 19 & 24 & 13 & 2\\ \ \end{array} \right] \bmod{26}$

$\implies$

$P^{-1} = \begin{bmatrix}{1} && {8} && {16} && {8} && {3} \\ {22} && {23} && {25} && {17} && {6} \\ {25} && {24} && {16} && {18} && {4} \\ {15} && {17} && {2} && {6} && {20} \\ {20} && {19} && {24} && {13} && {2}\\ \end{bmatrix} \bmod{26}$

$$

and $C = \begin{bmatrix}{15} && {7} && {24} && {0} && {12}\\ {7} && {0} && {11} && {13} && {8}\\ {14} && {19} && {4} && {14} && {13}\\ {18} && {8} && {19} && {11} && {4}\\ {15} && {3} && {7} && {0 } && {18}\\ \end{bmatrix} \bmod{26}$

$\implies$

$A = C * P^{-1} = \begin{bmatrix}{21} && {19} && {21} && {21} && {25} \\ {13} && {17} && {12} && {20} && {3} \\ {14} && {12} && {11} && {6} && {10} \\ {4} && {7} && {0} && {12} && {16} \\ {18} && {23} && {1} && {11} && {23}\end{bmatrix} \bmod{26}$ ,

$\implies \det(A) = \boxed{1}$ modulo 26.

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Incidentally, these are the only two 25 letter words in the English language that I could find.