problem solving blog #4: The Test of Bravery

A lot of times, the most difficult-looking problems have the easiest solutions, and the most simple-looking problems have the most difficult solutions.

Now, this problem looks simple at first, becomes difficult when you think about it, and if you’re brave enough, you can resolve it immediately.

Putnam and Beyond, #96

Have a think; be bold and be brave.

Photo by Mathew Schwartz on Unsplash

Honestly, when I first saw this, I thought it was going to require algebraic “intuition”. I felt really tempted to use Desmos, but I stopped myself from doing so.

If we are brave, can we make a bold guess? Of course, when n = 1, we can quickly see its real root at x = -1. Here’s how I proceeded to guess, which worked the first time:

1. Consider the polynomial corresponding to this equation. The leading coefficient is 1, and the degree of the polynomial is 4; that means, while the latter terms have a chance of exceeding the leading term for lower values, this becomes impossible as the absolute value of x increases.
2. There is a positive constant term to balance things out for smaller values, hence there is a high likelihood that the polynomial remains non-negative for all real x.

What does this give us? We guess that

Somehow I thought of this immediately, which is quite unusual considering how I normally perform on algebra problems. I really hoped that this was true, because it would imply the following for n ≥ 2:

The inequality is strict, meaning that there’s no real roots in such cases.

Back to proving our claim. When encountering simplistic polynomial expressions, it is usually a good idea to express the polynomial as sum of squares; that way, it will always be nonnegative.

That was the first thing I did. Starting with

But, then I realized we needed more square terms. The above expression generates a positive square term; then the next expression would need to generate a negative square term, which is unideal, since the degree of the next expression should not exceed 2. Hence, I corrected this to

The remaining factorization of quadratic terms is simple;

After that,

which has unique equality at x = -1. Now we can easily show that the only possible value of n is 1.

So, in summary

You have to make bold guesses in math and believe in your guesses firmly, even if you aren’t sure if they’re correct.

That’s it! It’s that simple.