Elegant route

I dabble in mathematics at a puzzle-solving level. Although I took it as a first-year subject at university, I never had that extra spark of aptitude that makes a mathematician. However, very very rarely I find problems that briefly touch what I imagine real mathematics to be.

Recently I encountered this problem:

Solve (x+2)(x+4)(x+6)(x+8) = 9

Now the temptation is to charge straight at it, and I did, first expanding terms of the left hand side and rearranging to a standard quartic ...

(x+2)(x+4)(x+6)(x+8) = 9
(x^2 + 6x + 8)(x + 6)(x + 8) = 9
(x^3 + 12x^2 + 44x + 48)(x + 8) = 9
x^4 + 20x^3 + 140x^2 + 400x + 384 = 9
x^4 + 20x^3 + 140x^2 + 400x + 375 = 0

This proved to be horrible. A lot of slog through possible small-integer zeros found x = -5 to be a solution. This enabled a long division by (x+5) to give this factorisation:

(x + 5)(x^3 + 15x^2 + 65x + 75) = 0

But the remaining cubic seemed impossible to tackle without using a maths package.

But then I noticed the symmetry of (x+2)(x+4)(x+6)(x+8) around x+5, and I tried this substitution:

Let p = x + 5 ... so the problem becomes:

(p - 3)(p - 1)(p + 1)(p + 3) = 9

This leads nicely to a reverse application of the standard "difference of two squares" factorisation, and a greatly simplified result.

(p - 3)(p - 1)(p + 1)(p + 3) = 9
(p - 3)(p + 3)(p - 1)(p + 1) = 9 ... rearrange terms
(p^2 - 9)(p^2 - 1) = 9 ... expand to a pair of "difference of two squares"
p^4 - 10p^2 + 9 = 9 ... expand
p^4 - 10p^2 = 0 ... simplify

This is easy to solve for p:

p^4 - 10p^2 = 0
p^2(p^2 - 10) = 0
So p^2 = 0, or p^2 = 10
So p = 0, or p = √10 or p = -√10

And finally we can reverse the substitution: x = p - 5

So x = -5 or x = √10 - 5 or x = -√10 - 5

It's only a puzzle, but the elegance and clarity of the second solution route compared to the murky dead end of the routine one is oddly satisfying.

(The quartic has only three zeros, all real, by the way, because its central maximum at x = -5 touches the x-axis).

- Ray