Question

prove that if n is a perfect square then n + 2 is not a perfect square

Answer 1

Answer:

This statement can be proven by contradiction for $n \in \mathbb{N}$ (including the case where $n = 0$.)

$\text{Let n \in \mathbb{N} be a perfect square}$.

$\textbf{Case 1.} ~ \text{n = 0}$:

$\text{ n + 2 = 2 , which isn't a perfect square}$.

$\text{Claim verified for n = 0 }$.

$\textbf{Case 2.} ~ \text{ n \in \mathbb{N} and n \ne 0 . Hence n \ge 1 }$.

$\text{Assume that n is a perfect square}$.

$\text{ \iff \exists a \in \mathbb{N} s.t. a^2 = n }$.

$\text{Assume \textit{by contradiction} that (n + 2) is a perfect square}$.

$\text{ \iff \exists b \in \mathbb{N} s.t. b^2 = n + 2 }$.

$\text{ n + 2 > n > 0 \implies b = \sqrt{n + 2} > \sqrt{n} = a }$.

$\text{ a,\, b \in \mathbb{N} \subset \mathbb{Z} \implies b - a = b + (- a) \in \mathbb{Z} }$.

$\text{ b > a \implies b - a > 0 . Therefore, b - a \ge 1 }$.

$\text{ \implies b \ge a + 1 }$.

$\text{ \implies n+ 2 = b^2 \ge (a + 1)^2= a^2 + 2\, a + 1 = n + 2\, a + 1 }$.

$\text{ \iff 1 \ge 2\,a }$.

$\text{ \displaystyle \iff a \le \frac{1}{2} }$.

$\text{Contradiction (with the assumption that a \ge 1 )}$.

$\text{Hence the original claim is verified for n \in \mathbb{N}\backslash\{0\} }$.

$\text{Hence the claim is true for all n \in \mathbb{N} }$.

Step-by-step explanation:

Assume that the natural number $n \in \mathbb{N}$ is a perfect square. Then, (by the definition of perfect squares) there should exist a natural number $a$ ($a \in \mathbb{N}$) such that $a^2 = n$.

Assume by contradiction that $n + 2$ is indeed a perfect square. Then there should exist another natural number $b \in \mathbb{N}$ such that $b^2 = (n + 2)$.

Note, that since $(n + 2) > n \ge 0$, $\sqrt{n + 2} > \sqrt{n}$. Since $b = \sqrt{n + 2}$ while $a = \sqrt{n}$, one can conclude that $b > a$.

Keep in mind that both $a$ and $b$ are natural numbers. The minimum separation between two natural numbers is $1$. In other words, if $b > a$, then it must be true that $b \ge a + 1$.

Take the square of both sides, and the inequality should still be true. (To do so, start by multiplying both sides by $(a + 1)$ and use the fact that $b \ge a + 1$ to make the left-hand side $b^2$.)

$b^2 \ge (a + 1)^2$.

Expand the right-hand side using the binomial theorem:

$(a + 1)^2 = a^2 + 2\,a + 1$.

$b^2 \ge a^2 + 2\,a + 1$.

However, recall that it was assumed that $a^2 = n$ and $b^2 = n + 2$. Therefore,

$\underbrace{b^2}_{=n + 2)} \ge \underbrace{a^2}_{=n} + 2\,a + 1$.

$n + 2 \ge n + 2\, a + 1$.

Subtract $n + 1$ from both sides of the inequality:

$1 \ge 2\, a$.

$\displaystyle a \le \frac{1}{2} = 0.5$.

Recall that $a$ was assumed to be a natural number. In other words, $a \ge 0$ and $a$ must be an integer. Hence, the only possible value of $a$ would be $0$.

Since $a$ could be equal $0$, there's not yet a valid contradiction. To produce the contradiction and complete the proof, it would be necessary to show that $a = 0$ just won't work as in the assumption.

If indeed $a = 0$, then $n = a^2 = 0$. $n + 2 = 2$, which isn't a perfect square. That contradicts the assumption that if $n = 0$ is a perfect square, $n + 2 = 2$ would be a perfect square. Hence, by contradiction, one can conclude that

$\text{if n is a perfect square, then n + 2 is not a perfect square.}$.

Note that to produce a more well-rounded proof, it would likely be helpful to go back to the beginning of the proof, and show that $n \ne 0$. Then one can assume without loss of generality that $n \ne 0$. In that case, the fact that $\displaystyle a \le \frac{1}{2}$ is good enough to count as a contradiction.