Solving Problems From The Back

Most of you have heard about "working backwards", where you start at the end and work your way through equations to figure out what the starting conditions are. When it comes problem solving, there is a similar approach of working backwards that allows you to figure out how to find a possible path towards your end goal.

I am a huge champion / believer in "working from the front and from the back to see how they meet in the middle". (Yes I need a much shorter name for that, any suggestions?) Let me explain that with an interesting problem that I recently came across:

India Selection Test
Find all such polynomial f(x)f(x) with integer coefficients such that for all integers nn, we have gcd⁡(f(n),f(2n))=1 \gcd(f(n),f(2^n))=1.

At first glance, it is hard to see how one can start to prove this statement. Who am I kidding, at second or third glances, there is no clear path forward. This is when it helps to "work from the end result", and see what hints (breadcrumbs) are left for us to follow.

We will use the well known results that
1) For any f∈Z[x] f \in \mathbb{Z} [x], a−b∣f(a)−f(b) a-b \mid f(a) - f(b).
2) Fermat's Little Theorem (Euler's Theorem)
Exercise 1: Prove these results.

Hint: Read Polynomial Sprint: Useful Lemma.

Ponder this, and then move on to the next note in this set.

#Algebra #ProofTechniques

Note by Calvin Lin
6 years, 11 months ago

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