You may also like

problem icon

Real(ly) Numbers

If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?

problem icon

Roots and Coefficients

If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?

problem icon

Pair Squares

The sum of any two of the numbers 2, 34 and 47 is a perfect square. Choose three square numbers and find sets of three integers with this property. Generalise to four integers.

Quadratic Harmony

Stage: 5 Challenge Level: Challenge Level:1

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, $a\geq b$.

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).