Bertie

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 #5
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Unless we are told something to the contrary, unknowns in equations such as this are always assumed to be integers.

 

If  d is the greatest common divisor (gcd) of a and n, the linear congruence  

$$\displaystyle ax\equiv b \bmod n$$ 

has a solution only if b is also divisible by d and in that case there will be an infinite number of solutions given by

$$\displaystyle x=x_{0}+\frac{nt}{d},$$

where  $$x_{0}$$  is any solution whatever, and t is an integer, (positive or negative), or zero.

For the equation in question,

$$\displaystyle 7x = 1\bmod 26,\qquad d = 1,$$

so the general solution will be

$$\displaystyle x = x_{0}+\frac{26t}{1} = x_{0}+26t.$$

Finding a suitable  $$x_{0}$$  is a trial and error thing, run through the multiples of 7 and 26 looking for the first pair for which the multiple of 7 is 1 greater than the multiple of 26 That happens to be 15 and 4, so the general solution is

$$\displaystyle x = 15 + 26t, \quad t\:\in \:Z.$$ 

 

The result given earlier comes from the theory of linear Diophantine equations.

If

$$\displaystyle d = \gcd(a,b),$$

the equation

$$\displaystyle ax+by = c$$

has integer solutions only if c is also divisible by d, and in that case there will be an infinite number of solutions given by

$$\displaystyle x = x_{0}+\frac{bt}{d},\quad y=y_{0}-\frac{at}{d} \quad (t\: \in Z),$$

and where  $$\displaystyle \; x_{0},y_{0}$$  is any particular solution.

(For large values of a and b, $$\displaystyle \; x_{0},y_{0}$$  and d are usually found using Euclid's algorithm.)

( The equation

$$\displaystyle ax\equiv b \bmod n$$

is equivalent to

$$\displaystyle ax-b=kn \Rightarrow ax-kn=b,$$

which, with a change of letters and a change of sign, is the Diophantine equation above.)

May 8, 2015