Mathematics: Prove that if n is a perfect square then n + 2 is not a perfect square

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Prove that if n is a perfect square then n + 2 is

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TekoaDaHedgie12

15.01.2022

Mathematics

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 ( $a \in \mathbb{N}$ ) 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 and are natural numbers. The minimum separation between two natural numbers is . 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 was assumed to be a natural number. In other words, $a \ge 0$ and must be an integer. Hence, the only possible value of would be .

Since could be equal , 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, 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.

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

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