2004 Indonesia MO Problems/Problem 6

Problem

A quadratic equation $x^2 + ax + b + 1 = 0$ with integers $a$ and $b$ has roots which are positive integers. Prove that $a^2 + b^2$ is not a prime.

Solution

Let $r$ and $s$ be the two positive roots of the quadratic. By Vieta's Formulas, \[r + s = -a\] \[rs-1 = b\] Square both equations to get \[r^2 + 2rs + s^2 = a^2\] \[r^2s^2 - 2rs + 1 = b^2\] Add the two equations to get \[a^2 + b^2 = r^2 + 2rs + s^2 + r^2s^2 - 2rs + 1\] \[a^2 + b^2 = r^2 + s^2 + r^2s^2 + 1\] The right hand side can be factored into $(r^2 + 1)(s^2 + 1)$. Since $r,s > 0$, $r^2 + 1 > 1$ and $s^2 + 1 > 1$, so $a^2 + b^2$ can never be a prime since it can be written as the product of two integers greater than one.

See Also

2004 Indonesia MO (Problems)
Preceded by
Problem 5
1 2 3 4 5 6 7 8 Followed by
Problem 7
All Indonesia MO Problems and Solutions