The question of whether the expression $n^2 + 4n$ can be a perfect square for integer values of $n$ is a classic problem in elementary number theory. It might seem like a simple problem at first glance, but it requires a bit of clever manipulation and understanding of number properties to arrive at a definitive answer. Let's delve into the problem and explore its solution.
Exploring the Problem
To determine if $n^2 + 4n$ can be a perfect square, we need to investigate whether there exists an integer $k$ such that:
$n^2 + 4n = k^2$
We can rewrite the equation as follows:
$k^2 - n^2 = 4n$
Factoring the left side, we get:
$(k-n)(k+n) = 4n$
This equation tells us that the product of two integers, $(k-n)$ and $(k+n)$, must be equal to four times another integer, $n$. To analyze this further, let's consider the possible factors of $4n$.
Case Analysis and Solution
Let's analyze the factors of $4n$ and the corresponding values of $(k-n)$ and $(k+n)$:
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Case 1: Both factors are even
- If both $(k-n)$ and $(k+n)$ are even, then their product is divisible by 4. However, this also means that $n$ must be even. Let's express $n$ as $2m$ where $m$ is an integer. Substituting into the equation, we get:
$(k-2m)(k+2m) = 8m$
We can further simplify this to:
$k^2 - 4m^2 = 8m$
This equation implies that $k^2$ must be even. Since the square of an odd number is odd, $k$ must also be even. Let's express $k$ as $2p$ where $p$ is an integer. Substituting into the equation, we get:
$4p^2 - 4m^2 = 8m$
Dividing both sides by 4, we get:
$p^2 - m^2 = 2m$
We can rewrite this equation as:
$(p-m)(p+m) = 2m$
This equation implies that $m$ must be even (since the product of two consecutive integers is always even). This leads to a contradiction because we initially assumed $m$ to be any integer. Therefore, both factors cannot be even.
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Case 2: One factor is even, the other is odd
- If one factor is even and the other is odd, their product is even. This means that $n$ must be even. Let's express $n$ as $2m$ where $m$ is an integer. Substituting into the equation, we get:
$(k-2m)(k+2m) = 8m$
This equation implies that $k$ must be even. Let's express $k$ as $2p$ where $p$ is an integer. Substituting into the equation, we get:
$4p^2 - 4m^2 = 8m$
Dividing both sides by 4, we get:
$p^2 - m^2 = 2m$
This equation leads to a contradiction, as we discussed in Case 1. Therefore, one factor cannot be even and the other odd.
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Case 3: Both factors are odd
- If both $(k-n)$ and $(k+n)$ are odd, then their product is odd. However, this means that $4n$ must be odd. This is impossible since $4n$ is always even. Therefore, both factors cannot be odd.
Since all possible cases lead to contradictions, we can conclude that there is no integer value of $n$ that can make $n^2 + 4n$ a perfect square.
Conclusion
By analyzing the factors of $4n$ and considering different cases, we have definitively shown that the expression $n^2 + 4n$ cannot be a perfect square for any integer value of $n$. This simple problem demonstrates the power of factorization and careful case analysis in solving problems in number theory.
