# Prove that if u is an odd integer, then the equation x^2+x-u=0 has no solution that is an integer?

## This question comes from a section on contradiction and proof by cases. I thought that I could use the logical equivalency $p \implies q \equiv \neg q \implies \neg p$, but I'm sure how to go about showing that something has an integer solution either in the case of this particular equation. Any guidance would be much appreciated. (This question is from Mathematical Reasoning by Ted Sundstrom, Sect. 3.4 Q 2)

Feb 17, 2017

Hint 1: Suppose that he equation ${x}^{2} + x - u = 0$ with $u$ an integer has integer solution $n$. Show that $u$ is even.

#### Explanation:

If $n$ is a solution the there is an integer $m$ such that

${x}^{2} + x - u = \left(x - n\right) \left(x + m\right)$

Where $n m = u$ and $m - n = 1$

But the second equation entails that $m = n + 1$

Now, both $m$ and $n$ are integers, so one of $n$, $n + 1$ is even and $n m = u$ is even.

Feb 17, 2017

Proposition
If $u$ is an odd integer, then the equation ${x}^{2} + x - u = 0$ has no solution that is an integer.

Proof
Suppose that there exists a integer solution $m$ of the equation:

${x}^{2} + x - u = 0$

where $u$ is an odd integer. We must examine the two possible cases:

$m$ is odd; or
$m$ is even.

First, let us consider the case where $m$ is odd, then there exists an integer $k$ such that:

$m = 2 k + 1$

Now, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
 :. (2k + 1)^2 + (2k + 1) − u = 0
 :. (4k^2 + 4k + 1) + (2k + 1) − u = 0
 :. 4k^2 + 6k + 2 − u = 0
$\therefore u = 4 {k}^{2} + 6 k + 2$
$\therefore u = 2 \left(2 {k}^{2} + 3 k + 1\right)$

And we have a contradiction, as $2 \left(2 {k}^{2} + 3 k + 1\right)$ is even, but $u$ is odd.

Next, let us consider the case where $m$ is even, then there exists an integer $k$ such that:

$m = 2 k$

Similarly, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
 :. (2k)^2 + (2k) − u = 0
 :. 4k^2 + 2k − u = 0
$\therefore u = 4 {k}^{2} + 2 k$
$\therefore u = 2 \left(2 {k}^{2} + k\right)$

And, again, we have a contradiction, as $2 \left(2 {k}^{2} + k\right)$ is even, but $u$ is odd.

So we have proved that there is no integer solution of the equation ${x}^{2} + x - u = 0$ where $u$ is an odd integer.

Hence the proposition is proved. QED

May 12, 2017

See below.

#### Explanation:

If ${x}^{2} + x - u = 0$ then

$x \left(x + 1\right) = u$ then if $x$ is an integer, $x \left(x + 1\right)$ is even, being a contradiction because $u$ by hypothesis is odd.