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: Proof by McWorter)
Supposing that n=ab, and a and b are coprime, then ac-bd=+-1 for
positive integers c,d. By the theorem that Ax-By=1 is solvable for the
integers x,y if and only if
A and B are coprime, if x=X, y=Y is a solution, then X and B are
coprime (as are X and Y, and A and Y). Hence c cannot be a multiple of
b nor can d be a multiple of a. Hence ac and bd are consecutive
numbers, and neither is a multiple of n=ab. Hence ac and bd are an x-1
and an x-2 in some order for an x that is neither 1 nor 2 mod n=ab.
Fun problem, McW!
Edited on September 3, 2005, 11:37 pm
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Posted by Richard
on 2005-09-03 23:09:09 |