Congratulations John from State College
Area High School, Pennsylvania, USA, Andrei from School 205,
Bucharest, Romania and Marcos from Cyprus on your excellent
solutions to Quadratic Harmony.
Say that x2-a x+b=0 has roots a, b. Then a+b = a and
ab = b. Without loss of generality, a ³ b.
Case 1: a=b. Then x2-a x+a = 0. So a+b = ab, so
ab-a-b = 0, so (a-1)(b-1)=1. Since a and
b are natural numbers, we must have a-1=1 and b-1=1, so
a = 2=b, so a=4=b.
Case 2: a > b. Then a+b > ab, so (a-1)(b-1) < 1, so
(a-1)(b-1)=0, so a = 1 or b = 1. Without loss of generality,
a = 1, so b=b and a=b+1=b+1. So the quadratics are x2-(b+1)x+b
and x2-b x+b+1. The first of these has roots 1 and b, as we expected. So
we just need the second one to have natural number roots. So certainly b2-4b-4
(the discriminant) is a square, say b2-4b-4=X2. Then (b-2-X)(b-2+X)=8.
We can quickly check that we can't have 8, 1 as this gives a value of b that
isn't an integer. So we have b-2+X=4, b-2-X=2, so b=5. Now we are trying
to solve x2-6x+5=0 and x2-5x+6=0, and these are clearly both soluble in
positive integers.
So, to summarise, the only possible values are a=4, b=4, and a=5, b=6
(or obviously a and b reversed).