The quadratic equation x^2-3x+2=0 has the "correct" number of solutions modulo 5 and 7. However, modulo 6 the equation has four solutions; namely, 1, 2, 4, and 5. For what positive integers n does the equation x^2-3x+2=0 have exactly two incongruent solutions modulo n?

(In reply to re(5): Proof by McWorter)

I really don't understand what is not clear. Given coprime a and b, there is a pair of numbers, namely ac and bd, neither of which is congruent to 0 mod ab, and which differ from one another by 1 so we can set one of them equal to x-1 and the other equal to x-2, for some x. Hence neither x-1 nor x-2 is congruent to 0 mod ab, so that x is not congruent to either 1 or 2 mod ab. For example, if x were congruent to 1, then x-1 would necessarily be congruent to 0, which it is not since it is equal to ac which has been proved to not be congruent to 0, all congruences being mod ab.

The product (ac)(bd)=(ab)(cd)=(x-1)(x-2) is clearly congruent to 0 mod ab, and x is congruent to something other than 1 or 2 mod ab=n, then. So this x is another root of the equation besides the two roots 1 and 2 that always clearly exist regardless of the value of the modulus n.

Perhaps there is some confusion because of the uncertainty of which of ac or bd is actually the greater. What we can do is make the proviso that ac is the greater by insisting that we have chosen between two given coprime factors of n in such a way that the one we call a is the one that makes ac greater than bd. Then x-1=ac and x-2=bd and x=1+ac=2+bd.

 

 

 

 

 

 

Edited on September 4, 2005, 5:09 am

Posted by Richard on 2005-09-04 04:38:50