Consider the following problem: The square of the difference of two numbers is the same as their sum. What numbers satisfy this condition?
We might begin by expressing the problem as an equation: \((a - b)^2 = a + b\). Then we can look for possible solutions. Let's try to simplify the expression:
$$ \begin {eqnarray} a^2 - 2ab + b^2 &=& a + b \\ a^2 - 2ab + b^2 - a - b &=& 0 \\ a^2 - (2b + 1)a + (b^2 - b) &=& 0 \end {eqnarray} $$Solving for \(a\) yields:
$$ \begin {eqnarray} a &=& {{(2b + 1) \pm \sqrt{(2b + 1)^2 - 4(b^2 - b)}} \over 2} \\ a &=& {{(2b + 1) \pm \sqrt{(4b^2 + 4b + 1) - 4b^2 + 4b}} \over 2} \\ a &=& {{(2b + 1) \pm \sqrt{8b + 1}} \over 2} \end {eqnarray} $$This last equation has an integer solution whenever \(8b + 1\) is a perfect square. This is so because as long as \(b\) is an integer, both \((2b + 1)\) and \((8b + 1)\) will be odd. Their sum or difference will be even, so \(a\) will be a whole number. So all we need to do is find values for \(b\) that make \(8b + 1\) a perfect square. For example, if \(b = 1\), then \(8b + 1 = 9\) is a perfect square; or if \(b = 3\), then \(8b + 1 = 25\) is a perfect square. But if \(b = 5\), then \(8b + 1 = 41\) is not a perfect square. This is still a hit or miss way of solving the problem. We have two solutions:
$$ \begin {eqnarray} b = 1 \rightarrow a &=& {{(2 \times 1 + 1) \pm \sqrt{8 \times 1 + 1}} \over 2} \\ a &=& {{3 \pm 3} \over 2} \\ a &=& 0, 3 \end {eqnarray} $$This leads to the pair of solutions: \((0 - 1)^2 = (0 + 1)\) and \((3 - 1)^2 = (3 + 1)\). We also have:
$$ \begin {eqnarray} b = 3 \rightarrow a &=& {{(2 \times 3 + 1) \pm \sqrt{8 \times 3 + 1}} \over 2} \\ a &=& {{7 \pm 5} \over 2} \\ a &=& 1, 6 \end {eqnarray} $$This leads to the pair of solutions: \((1 - 3)^2 = (1 + 3)\) and \((6 - 3)^2 = (6 + 3)\).
We would like to develop a more general solution, though: one that does not involve trial and error. Let's try a slightly different approach to the question of finding perfect squares that can be expressed as \(8b + 1\). If we divide \(8b + 1\) by 8 for any whole number \(b\), the remainder will always be 1.
So let's ask this question: What perfect squares give a remainder of 1 when divided by 8? We can start with a list of perfect squares and their remainders:
|
Square |
Remainder |
|---|---|
|
1 |
1 |
|
4 |
4 |
|
9 |
1 |
|
16 |
0 |
|
25 |
1 |
|
36 |
4 |
|
49 |
1 |
|
64 |
0 |
|
81 |
1 |
It certainly looks as though every odd square gives a remainder of 1 when divided by 8. Consider the odd square \((2n - 1)^2\): $$ \begin {eqnarray} (2n - 1)^2 &=& 4n^2 - 4n + 1 \\ &=& 4n(n - 1) + 1 \end {eqnarray} $$
Clearly if \(n\) is even, then \(4n(n - 1)\) is divisible by \(8\). Likewise, if \(n\) is odd, then \((n - 1)\) is even, and again \(4n(n - 1)\) is divisible by 8. Therefore, whether \(n\) is even or odd, \((2n - 1)^2\) has a remainder of 1 when divided by 8. We now have a method for generating solutions. For every odd square, subtract \(1\) and divide by \(8\) to get \(b\) Then use the value of \(b\) to obtain the value of \(a\) Using the equation above, we can express \(b\) as \({{[4n(n - 1) + 1] - 1} \over 8}\). This simplifies to \(b = {{n(n - 1)} \over 2}\). This works for any whole number \(n\). We can substitute this value to solve for \(a\).
$$ \begin {eqnarray} a &=& {{2\left({n(n - 1) \over 2}\right)} + 1 \pm \sqrt{8\left({n(n - 1) \over 2}\right) + 1} \over 2} \\ &=& {{n^2 - n + 1 \pm (2n - 1)} \over 2} \\ &=& {{n^2 + n} \over 2}, {{n^2 - 3n - 2} \over 2} \\ a &=& {n(n + 1) \over 2}, {(n - 1)(n - 2) \over 2} \end {eqnarray} $$For every number \(n\), there exist two other numbers \(a = {n(n + 1) \over 2}\) and \(b = {n(n - 1) \over 2}\) such that \((a - b)^2 = a + b\). If these numbers look familiar, it is because they are triangle numbers. A triangle number has the form \(T_{n} = {n(n + 1) \over 2}\). It should be apparent that \(T_{n - 1} = {n(n - 1) \over 2}\) and \(T_{n - 2} = {(n - 1)(n - 2) \over 2}\). For any two adjacent triangle numbers \(T_{n - 1}\) and \(T_{n}\) their difference \(T_{n} - T_{n - 1} = n\) and their sum \(T_{n} + T_{n - 1} = n^2\). So any two adjacent triangle numbers will satisfy the original conditions of the problem. For example, \((21 - 15)^2 = 21 + 15\).