How To Really Solve This Tricky Algebra Problem?

On the surface, this does not look like a tricky algebra problem. You have an equation where one sum equates to another sum. All the terms have the same integer exponent (x). Your job is just to figure out the integer value(s) for which this equation holds.

As you continue reading this essay, I will first demonstrate an easy way to come up with one sample solution for the challenge. After this, I will proceed to explain a couple of finer mathematical concepts that are useful to explore further solutions (if any).

Spoiler Alert:

If you wish to try this problem out on your own, now is the time. Once you are done with your version of the solution, you may choose to continue reading. From this point on, I will be explicitly discussing the solution(s) to the problem.

This essay is supported by Generatebg

The First Sample Solution

Coming up with the first sample solution is not all that difficult. We could just choose to use the plain old substitution method. Let us first define the left-hand side of the equation as L.H.S and the right-hand side of the equation as R.H.S. When I start substituting integer values from zero upwards (along the positive number line), I get the following result:

It looks like we have arrived at a solution just on our third attempt! The integer 2 is a valid exponent that holds for this equation. Now, the tricky part of the challenge begins.

How are we going to figure out if there are other possible integer solutions?

There could be positive integers that are greater than 2 or negative integers that could be potential solutions as well. We could continue substituting values, but that would simply be brute-forcing the solutions.

Thankfully, there are more elegant approaches to such problems in mathematics. We will look at one such approach. But before we proceed with the approach, there are a couple of fundamental concepts that we’ll need to cover first.

---

Increasing Functions and Decreasing Functions

Whenever we deal with functions in mathematics, certain problems (like our challenge here) could be solved faster and more efficiently by analysing the properties of the function we are dealing with.

“But wait a minute! There are no explicit functions in the main problem statement. How is function analysis going to help us?”

If you have concerns along these lines, worry not. We will eventually get to that part. Moving back to properties of functions, one useful property is whether the function is increasing or decreasing.

Increasing and decreasing functions are defined as follows:

Furthermore, strictly increasing and decreasing functions are defined as follows:

Below, you can find graphical illustrations of each of these function types (but please note that these are hand-drawn illustrations only; not mathematical perfections):

Now that we have the necessary concepts, we may proceed to solve the original problem.

---

Solution to the Tricky Algebra Problem

Sofar, we know that integer 2 is definitely a valid solution. Now, to avoid brute-force substitution, I divide the entire equation by 13^x. Consequently, I get the following as the result:

Now, take a look at the L.H.S and R.H.S of this transformed equation. It is not quite apparent, but the L.H.S is a strictly decreasing function, whereas the R.H.S is a strictly increasing function. Let me make it a bit more apparent by plotting integer values from one to ten for x for each case.

There is something unique about an equation whose L.H.S is strictly decreasing and R.H.S is strictly increasing. Such an equation can have only one intersection point. If this is hard for you to visualise, it becomes apparent when we plot both L.H.S and R.H.S against integer x-values on the same plot as follows:

Therefore, we can say with certainty that there are no other integer solutions for the given problem other than 2 (refer to the appendix below for a rigorous treatment).

---

Appendix:

We may choose to compute the individual terms on the L.H.S and the R.H.S of the transformed equation. When we do this, we see that all the base terms on the L.H.S are less than 1 and the only base term on the R.H.S is greater than 1. We can now apply mathematical induction (you can learn more about mathematical induction from my essay on this topic). When we do that, we see that the L.H.S is strictly decreasing and the R.H.S is strictly increasing.

---

Credit: Andreescu T., and Feng Z.

If you’d like to get notified when interesting content gets published here, consider subscribing.

Further reading that might interest you: How To Really Deal With The Friendship Paradox? and The Shipwreck Puzzle: A Fun Linear Math Challenge.

If you would like to support me as an author, consider contributing on Patreon.