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Modular Numbers and Cryptography

Modular Equations

Solve \( (4 + x) \equiv 5 \pmod{7}\)

A modular system \pmod{n} allows only a fixed set of remainder values, \(0,1,2,\ldots,n-1\). One practical approach to solving modular equations, at least when n is reasonably small, is to simply try all these integers. For each solution found, other can be found by adding multiples of the modulus to it.

Because dividing 5 by 7 yields a remainder 5, the criterion for congruence is that the given equation is true only if dividing \(4+x\) by 7 also yields remainder 5. In other words, find all values of x such that, a remainder of 5 is the result when \(4+x\) is divided by 7.

Since modulo 7 has residues \(0,1,2,3,4,5,6\), let's test each of them.

\( x=0: (4+0) \equiv 5 \pmod{7}\) is false. The remainder is 4.

\( x=1: (4+1) \equiv 5 \pmod{7}\) is true. The remainder is 5.

\( x=2: (4+2) \equiv 5 \pmod{7}\) is false. The remainder is 6.

\( x=3: (4+3) \equiv 5 \pmod{7}\) is false. The remainder is 0.

\( x=4: (4+4) \equiv 5 \pmod{7}\) is false. The remainder is 1.

\( x=5: (4+5) \equiv 5 \pmod{7}\) is false. The remainder is 2.

\( x=6: (4+6) \equiv 5 \pmod{7}\) is false. The remainder is 3.

\(\therefore x =1 \)  is a solution of the equation \( (4 + x) \equiv 5 \pmod{7}\). We can find other solutions by adding the modulus 7.

\[1+7=8, 8+7=15, 15+7=22, \ldots\]

The set of all nonnegative solutions of \( (4 + x) \equiv 5 \pmod{7}\) is \(\{1,8,15,22,29,\ldots\}\) or \(1+7n,n\in\mathbb{Z}\). There are infinite many solutions stemming from the residue 1 in \(\mod{7}\). However, there can be more than one residue as the solution.


Solve \( (3x + 2) \equiv 5 \pmod{9}\)

When we test the residues we find,

\( x=1: (3[1]+2) \equiv 5 \pmod{9}\) is true. The remainder is 5.

\( x=4: (3[4]+2) \equiv 5 \pmod{9}\) is true. The remainder is 5.

\( x=7: (3[7]+2) \equiv 5 \pmod{9}\) is true. The remainder is 5.

So the solution is when \(x = 1+9n,4+9n,7+9n, n\in\mathbb{Z}\).


Solve \(2x \equiv 3 \pmod{6} \)

When we test all the residues, none of them will have a remainder of 3, that is because every value gets doubled and becomes an even number. Thus, there is no solution.

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