Background

I was working through the practice problems found in College Algebra Seventh Edition by Stewart, Redlin, and Watson when I stumbled upon this problem found in 1.4 Exercises, Question 68. The question is stated as:

The sum of the squares of two consecutive even integers is 1252. Find the integers.

I found this question very enjoyable as:

  1. You needed to have a concrete definition of an even integer
  2. Solve the equation from your knowledge of quadratic equations

Solving the equation

We can formally state the question as

$$ m^2 + n^2 = 1252\ \ and\ m, n \in \mathbb{Z} $$

where:

$m, n \in \mathbb{Z}$ , $m$ and $n$ are integers and

$m = n + 1$ , $m$ is the next consecutive integer after $n$

We can define an even integer as provided by Kenneth Rosen in his Discrete Mathematics books. An integer $n$ is even if for some integer $k$, $n = 2k$. We can intuitively see this by giving some examples of $k$.

When $k = { 1, 2, 3, 4, … }$ then $n = { 2, 4, 6, 8, … }$ and here we can see that for any $k$ if it is an even integer, it will be found in $n$.

How does this help us out? Well, now we can substitute values for $m$, and $n$ by the definition of an integer. We see that $m = 2k$ and $n = 2(k + 1) = 2k + 2$ where $k$ is the integer in question. Here we relate $m$ to $n$ as we know that $n$ must be the next integer in sequence after $m$.

Substituting our updated definition of $m$ and $n$ into the equation we get.

$$ m^2 + n^2 = 1252 $$ $$ (2k)^2 + (2k + 2)^2 = 1252 $$

Expanding the squares $$ 4k^2 + 4k^2 + 8k + 4 = 1252 $$

Simplifying $$ 8k^2 + 8k - 1248 = 0 $$ $$ k^2 + k - 156 = 0 $$

We can now solve the quadratic equation for $k$ using a multitude of tools, but we can use the quadratic formula for ease of computation. We see that the above simplified equation has the form $ax^2 + bx + c = 0$

Here: $a = 1,\ b = 1,\ c = -156$

Substituing into the quadratic formula:

$$ \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} $$ $$ k = \frac{-1\pm \sqrt{1^2 - 4(1)(-156)}}{2(1)} $$ $$ k = \frac{-1\pm \sqrt{1 + 624}}{2} $$ $$ k = \frac{-1\pm \sqrt{625}}{2} $$ $$ k = \frac{-1\pm 25}{2} $$ $$ \therefore\ k = 12,\ k = -13 $$

We have 2 solutions for $k$, and if we plug them into $m$, and $n$ we have.

For $k = 12$ $$ m = 2k = 2\times 12 = 24 $$ $$ n = 2k + 2 = 2\times 12 + 2 = 26 $$ $$ m^2 + n^2 = 24^2 + 26^ = 576 + 676 = 1252 $$

For $k = -13$ $$ m = 2k = 2\times -13 = -26 $$ $$ n = 2k + 2 = 2\times -13 + 2 = -24 $$ $$ m^2 + n^2 = 1252 = (-26)^2 + (-24)^2 = 676 + 575 = 1252 $$

From the above we can see that we have 2 solutions for the values of $m$ and $n$, that satisfies the sum of squares of two consecutive positive integers.

For the positive integer solution $$ m = 24,\ n = 26 $$

For the negative integer solution $$ m = -26,\ n = -24 $$

Disclaimer

None of this text has been created or proof read by any Generative AI. All of the content on this article has been manually typed by me.