A new December and a new bunch of puzzles from mscroggs.co.uk.

First let’s look at this landscape. For now ignore the green curve.

A solution p(x) has to fit between the blue and the red lines. It has to fit under the blue and over the red.

Could p(x) = a? No, a straight horizontal line wouldn’t be able to stay over the red line always.

Could p(x) = a + bx? Then p(0) = a, a positive integer. Then p(1) = a + b. The blue line goes through (1,1). And p(1) is an integer, at most 0. So this p(x) goes through (0,a), down to (1,a+b) and then up, up, up to stay above the red line. Not possible.

Could p(x) = a + bx + cx²? This is similar to the blue line, so maybe we choose p(x) = x² – 2x + 2 – a-little-bit. This would always be under the blue line. First guess: p(x) = x² – 2x + 1.

Is this above the red line? That’s equivalent to this being true:

4x – 9 < x² – 2x + 1 <=>

0 < x² – 6x + 10 <=>

0 < x² – 6x + 9 + 1 <=>

0 < (x – 3)² + 1

And yes, this is always true.

I haven’t checked whether this is the only solution.

p(23) = 23² – 2*23 + 1 = 484.

A closeup: