Suppose that p is prime and that a and b are positive integers with ( p , a ) =

Question

Suppose that p is prime and that a and b

are positive integers with (p, a) = 1. The following

method can be used to solve the linear congruence ax

≠¡ b (mod p).

Show that if the integer x is a solution of ax

≠¡ b (mod p), then x is

also a solution of the linear congruence

a1x

≠¡

–b[m/a] (mod p),

where a1 is the least

positive residue of p modulo a. Note that this

congruence is of the same type as the original congruence, with a

positive integer smaller than a as the coefficient of

x.

When the procedure of part (a) is iterated, one obtains a

sequence of linear congruences with coefficients of x equal

to a0 = a > a1 >

a2 >

 ·Ã‚ ·Ã‚ · . Show that there is

a positive integer n with an = 1,

so that at the nth stage, one obtains a linear congruence

x ≠¡ B (mod p).

Use the method described in part (b) to solve the linear

congruence 6x ≠¡ 7 (mod 23).

Explanation / Answer

This is not an easy question! As some other people said, using mod 2 can only take you as far as showing that you can't have more than one odd number out of the 4, that is, in the case where you have 2, 3 or 4 odd numbers, is trivial and you will get that a+b+c+d is not prime for 2 or 4 odd numbers, and that ab=cd is impossible if there are 3 odd numbers. The problem is with the case where only one of the 4 numbers is odd. A hard problem. I will show the simple version of the solution, and you will have to fill in the details! First, suppose that a=1. What will happen then? b = cd thus 1+ b + c + d = 1 + cd + c + d = 1 + c(d + 1) + d = (c + 1)(d + 1) which is not prime. Now, suppose that a = pq and that both p and q are primes: pqb = cd and therefore we can say (without loss of generality) that p divides d suppose that q does not divide d (if it does, things will be easier...) We get: b = cd/(pq) = (c/q)(d/p) and a + b + c + d = pq + (c/q)(d/p) + c + d = (c/q)(q+d/p) + pq + d = (c/q)(q+d/p) + p(q + d/p) = (p + c/q)(q + d/p) And thus, a + b + c + d is not prime. It is not hard to generalize this to any odd a, but I will leave this for you...

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