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

α

β

. Then

α + β = a

and

α β = b

. Without loss of generality,

a \geq b

Case 1: $a = b$ . Then $x^{2} - a x + a = 0$ . So $α + β = α β$ , so $α β - α - β = 0$ , so $(α - 1) (β - 1) = 1$ . Since $α$ and $β$ are natural numbers, we must have $α - 1 = 1$ and $β - 1 = 1$ , so $α = 2 = β$ , so $a = 4 = b$ .

Case 2: $a > b$ . Then $α + β > α β$ , so $(α - 1) (β - 1) < 1$ , so $(α - 1) (β - 1) = 0$ , so $α = 1 or β = 1$ . Without loss of generality, $α = 1$ , so $b = β$ and $a = β + 1 = b + 1$ . 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 $b^{2} - 4 b - 4$ (the discriminant) is a square, say $b^{2} - 4 b - 4 = X^{2}$ . 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 $x^{2} - 6 x + 5 = 0$ and $x^{2} - 5 x + 6 = 0$ , 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).