A random internet advertisement had a mathematical problem that intrigued me:

Solve for $x$

I found three different approaches to the problem. This equation had a bit of a hidden depth!

## Clever Solution

Presumably $x+15$ and $\mathrm{x}$ are both perfect squares. (Internet advertisements don't tend to deal with irrational numbers.) So we can assume that when we take the square roots, we will get integers back.

The easiest way to solve this problem is to use the fact that the summation of odd numbers results in the perfect squares.

$1=1$

$4=1+3$

$9=1+3+5$

$16=1+3+5+7$

...

There are other resources that go into it in more detail.

It is an interesting pattern that can be simplified as ${\left(n+1\right)}^{2}={n}^{2}+\left(2n+1\right)$. (As long as you note that $2n+1$ is an odd number.)

Since 15 is an odd number, then $x$ must be the perfect square that is 15 below another perfect square. With our simplified pattern equation above, we know that there must be a solution for: ${\left(n+1\right)}^{2}={n}^{2}+15$.

So $2n+1=15$. Trivially, $n=7$. Which means ${n}^{2}=49$.

So the answer is $x=49$. If we plug that into the equation, it works out.

## Mathematical Solution

Essentially, the easy solution is almost guessing as to what the answer is. This may not work as an answer if the problem was given by a pedantic math professor. So how would we mathematically prove it? With these steps:

$\sqrt{x+15}+\sqrt{\mathrm{x}}=15$ | Subtract $\sqrt{\mathrm{x}}$ from both sides. |

$\sqrt{x+15}=15-\sqrt{\mathrm{x}}$ | Square both sides |

$x+15=225-\mathrm{30}\sqrt{\mathrm{x}}+x$ | Simplify |

$\mathrm{30}\sqrt{\mathrm{x}}=210$ | Divide by 30 |

$\sqrt{\mathrm{x}}=7$ | Square both sides |

$\mathrm{x}=49$ | Solution! |

This matches our "easy solution" answer so we are happy.

## Curiosity

So, now that we’ve proven it beyond a shadow of a doubt, can we extrapolate it? Given a positive odd number $n$, is there an answer to every equation: $\sqrt{x+n}+\sqrt{x}=n$

Let's calculate and find out!

$\sqrt{x+n}+\sqrt{\mathrm{x}}=n$ | Subtract $\sqrt{\mathrm{x}}$ from both sides. |

$\sqrt{x+n}=n-\sqrt{\mathrm{x}}$ | Square both sides |

$x+n={n}^{2}-2\mathrm{n}\sqrt{\mathrm{x}}+x$ | Cancel the $x$ and rearrange |

$2\mathrm{n}\sqrt{\mathrm{x}}={n}^{2}-n$ | Divide by $n$ |

$2\sqrt{\mathrm{x}}=n-1$ | Divide by 2 |

$\sqrt{\mathrm{x}}=\frac{n-1}{2}$ | Square both sides (Possible because $\mathrm{n}\ge 1$) |

$\mathrm{x}={\left(\frac{n-1}{2}\right)}^{2}$ | Solution! |

And that is our solution: Given any odd number $n$ in the equation $\sqrt{x+n}+\sqrt{\mathrm{x}}=n$ the value of $x$ is ${\left(\frac{n-1}{2}\right)}^{2}$

This makes sense as $\frac{n-1}{2}$ is the inversion of our earlier equation for odd numbers: $2n+1$

If you look at the steps, the only assumption that is actually required is that $\mathrm{n}\ge 1$. So there is no requirement for $n$ to be an odd number. So the formula will work for even numbers like 2 as well:

However, the clean solutions with integers and perfect squares only works with positive odd numbers.

Thank you to Remkes Kooistra for checking my math

*2018-1-14*