Prove Non-Square Integer: Ceiling Function

[anemone]

Let $a$ and $b$ be two positive integers. Prove that the integer $ a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil$ is not a square.

(Here $\lceil z \rceil$ denotes the least integer greater than or equal to $z$.)

 

[jvicens]

Thank you for bringing up this interesting problem. I would like to provide a proof for the statement that $a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil$ is not a perfect square for any positive integers $a$ and $b$.

First, let us assume that $a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil$ is a perfect square, denoted by $c^2$, where $c$ is also a positive integer. This means that we can write the following equation:

$a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil = c^2$

We can rewrite the ceiling function as follows:

$a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil = a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil + 1-1 = a^2+1+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil -1$

Since $a^2$ and $1$ are perfect squares, their sum is also a perfect square. Therefore, we can rewrite the equation as:

$a^2+1+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil -1 = d^2$, where $d$ is a positive integer.

We can further simplify this equation by combining the terms $a^2$ and $1$:

$(a+1)^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil -1 = d^2$

Since $b$ is a positive integer, we can write $b \geq 1$. Therefore, $\dfrac{4a^2}{b} \geq 4a^2$. This means that the ceiling function will always round up to at least $4a^2$, making the term $\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil -1$ always greater than or equal to $4a^2-1$.

Substituting this into the equation, we get: