Use proof by contradiction to show that the equation $x^2+x+n = 0$ has no integer solution when n is odd.

Question

I am a little confused if my solution to the problem is correct. I tried to prove this by cases following a similar example where $n$ is negative, however I'm unsure if I'm correct in following this because in this problem, $n$ is positive.

Here's what I've tried to do so far:

Suppose that there exists an integer solution m of the equation $x^2 + x + n =0$ where n is an odd integer. We must observe 2 possible cases:

(i) m is odd and (ii) m is even

When m is odd, that is, there exists $j \in \mathbb{Z}$ such that $m = 2j + 1$.

Given that m is the root of the equation, we have $m^2+m+n = 0$

$\therefore (2j+1)^2 + (2j + 1) +n = 0$

$\therefore(4j^2+4j+1) + (2j+1)+n = 0$

$\therefore 4j^2 + 6j + 2 + n = 0$

$\therefore n = -4j^2 - 6k - 2$

$\therefore n = -2(2j^2 + 6j + 2)$

$\therefore n = -2k , k \in \mathbb{Z}$

This contradicts the hypothesis, since n is even.

I then followed a similar process for the second case, where I suppose that m is even, and I end my proof by concluding that there is no integer solution of the equation $x^2 + x + n = 0 $ when n is odd.

Is my process for this proof correct? Thank you very much in advanced for the help!

Answer 1

Yes, your proof is indeed correct. However, there is a much easier way to solve this problem.

$$x^2+x+n=0\implies x(x+1)=-n.$$

For $x\in \mathbb Z$, $x(x+1)$ is even and therefore, the equation has no integral solution.

Answer 2

If interested, here is another take on this question:

$$x^2 + x + n = \left(x + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{4} - n\right) = \left(x + \dfrac{1}{2} + \sqrt{\dfrac{1}{4} - n}\right) \left(x + \dfrac{1}{2} - \sqrt{\dfrac{1}{4} - n}\right) \\ = \left(x + \dfrac{1}{2} + \dfrac{\sqrt{1 - 4n}}{2}\right) \left(x + \dfrac{1}{2} - \dfrac{\sqrt{1 - 4n}}{2}\right), \text{ with }\space 1-4n > 0, \space n < \dfrac{1}{4} \implies n < 0$$

$$\left(\dfrac{1}{2} \pm \dfrac{\sqrt{1 - 4n}}{2}\right) \in \mathbb{Z} \implies 4n - 1 \text{ needs to be a perfect square }$$

$$\text{Assume odd $n$ satisfies $4n - 1 = a^2$. Any odd number can be expressed as $4k + 1 \lor 4k + 3$} \tag{1}$$

$$\begin{cases} (4k + 1)^2 = 16k^2 + 8k + 1 = 8(2k^2 + k) + 1 \implies (4k + 1)^2 \equiv 1 \mod{4} \\ \\ (4k + 3)^2 = 16 k^2 + 24k + 9 = 8(2k^2 + 3k + 1) + 1 \implies (4k + 3)^2 \equiv 1 \mod{4} \end{cases} \\ \quad \\ \text{Essentially,} \\ \quad \\ 4n - 1 = 4(n - 1) + 3 \\\implies -1 \equiv 3 \mod{4} \require{\cancel}\cancel{\large{\cong}} \begin{cases}1^2 \equiv 1 \mod{4} \\ 3^2 \equiv 1 \mod 4\end{cases} \\ \mathbf{\text{We have contradiction! }} \space \square$$

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From $(1)$ onwards, there is an easier way to prove $n$ must be even, but does not use contradiction: $$1 - 4n \equiv 1 \mod{8} \implies 2|{n}, \space n < 0$$