CTK Insights

23 Oct

An Equation in Rational Numbers

Just last week I received a request for a permission to use a picture from my Introduction to Graphs page in a forthcoming book. I gave the permission but noted that the picture is not mine and that I had lifted it from an old book years ago. None of the references listed on the page contained exactly that picture. Having no recollection where the picture came from, I began to search through my library. I have not found the picture yet, but did serendipitously come across several problems I have also forgotten.

Here's one such problem.

Find x, y, and z, that satisfy

x + y + z = x² + y² + z².

I made the formulation intentionally vague to enhance the effect of the coming clarification.

Quite obviously, in integers the only possible solutions are the combinations of 0s and 1s, like x = 1, y = z = 0, or y = 0, x = z = 1, etc. These solutions we may call trivial. Thus in integers the equation has only trivial solutions. However, this is not so is we allow x, y, z to be rational numbers. Furthermore, there is a really simple algorithm that reveals infinitely many solutions. And not only that. The algorithm has an obvious extension to similar equation in any number of variables, like

Find x, y, z, u, v, and w that satisfy

x + y + z + u + v + w = x² + y² + z² + u² + v² + w².

Solution →















Find x, y, and z, that satisfy

x + y + z = x² + y² + z².

Let a, b, c be arbitrary integers. Introduce

κ = (a + b + c) / (a² + b² + c²)

and let

x = κa, y = κb, z = κc.

Let's verify

x + y + z = κa + κb + κc = (a + b + c)² / (a² + b² + c²).

On the other hand,

x² + y² + z² = κ²(a² + b² + c²) = (a + b + c)²(a² + b² + c²) / (a² + b² + c²)²,

which is obviously the same.

For example, let a = 1, b = 2, c = 4. κ = 7 / 21 = 1/3. x = 1/3, y = 2/3, z = 4/3. It is easy to see that

1/3 + 2/3 + 4/3 = (1/3)² + (2/3)² + (4/3)².

The extension to any number of variables is indeed obvious. Further, we may reverse the process and ask for what κ the equation

κa + κb + κc = (κa)² + (κb)² + (κc)²

holds for some a, b, c. Obviously, the answer is the formula for κ listed above: κ = (a + b + c) / (a² + b² + c²). But then the next question comes out naturally: what about the solution to a modified equation

κa + κb + κc = (κa)³ + (κb)³ + (κc)³?

It looks like the equation x + y + z = x³ + y³ + z³ or even when passing to higher powers all have infinitely many solutions in rational numbers. Further extension includes mixing the powers, like in x² + y² + z² = x³ + y³ + z³.

Once you get the idea, the original equation x + y + z = x² + y² + z² does not look as intimidating and mysterious as it did at the outset.


  1. B. A. Kordemsky, Mathematical Temptations, Publishing House ONIKS, 2000 (in Russian)

One Response to “An Equation in Rational Numbers”

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