# Puzzles

## 10 December

The equation \(x^2+1512x+414720=0\) has two integer solutions.

Today's number is the number of (positive or negative) integers \(b\) such that \(x^2+bx+414720=0\) has two integer solutions.

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If the equation has two integer solutions, then it can be written as \((x+\alpha)(x+\beta)\), where \(\alpha\) and \(\beta\) are integers.
Expanding this and setting it equal to \(x^2+bx+414720=0\), we find that \(b=\alpha+\beta\) and \(414720=\alpha\beta\).

414720 has 55 different pairs of factors. Each of these pairs leads to two values of \(b\) (a positive and a negative value). Therefore there are **110** values of \(b\) that give the equation two integer solutions.

## Powerful quadratics

Find all real solutions to

$$(x^2-7x+11)^{(x^2-11x+30)}=1.$$

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If \(x^2-7x+11=1\) or \(x^2-11x+30=0\), then this is one. The solutions to these are \(x=2,5,\) and \(6\).

It could also be one if \(x^2-7x+11=-1\) and \(x^2-11x+30\) is even. This happens when \(x=3\) or \(4\).

Therefore all the solutions to this are \(x=2,3,4,5\) or \(6\).