Stage: 5 Challenge Level:
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 $x^2-a x+b=0$ has roots $\alpha$, $\beta$. Then
$\alpha+\beta=a$ and $\alpha\beta=b$. Without loss of generality,
Case 1: $a=b$. Then $x^2-a
x+a =0$. So $\alpha+\beta=\alpha\beta$, so
$\alpha\beta-\alpha-\beta=0$, so $(\alpha-1)(\beta-1)=1$. Since
$\alpha$ and $\beta$ are natural numbers, we must have $\alpha-1=1$
and $\beta-1=1$, so $\alpha=2=\beta$, so $a=4=b$.
Case 2: $a> b$. Then
$\alpha+\beta> \alpha\beta$, so $(\alpha-1)(\beta-1)< 1$, so
$(\alpha-1)(\beta-1)=0$, so $\alpha=1 or \beta=1$. Without loss of
generality, $\alpha=1$, so $b=\beta$ and $a=\beta+1=b+1$. So the
quadratics are $x^2-(b+1)x+b$ and $x^2-b x+b+1$. 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 $b^2-4b-4$ (the
discriminant) is a square, say $b^2-4b-4=X^2$. 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
$x^2-6x+5=0$ and $x^2-5x+6=0$, 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).