- Thread starter idontknow
- Start date

Here is the proof without solving the equations :

We can simply set \(\displaystyle x=(-1)^{t}\) and \(\displaystyle x-2=(-1)^{r}\) but it can be done easily below:

Since \(\displaystyle x^2 -2x = x(x-2) \) then \(\displaystyle -1\leq (x,x-2) \leq 1\) or \(\displaystyle x=(-1)^{t}\) and \(\displaystyle x-2=(-1)^{t}-2<0\).

\(\displaystyle x-2<0\) means \(\displaystyle (-1)^{t-x} <0\).

The only negative value \(\displaystyle (-1)^{t-x}\) can take is -1.(if it exists, from now there is one solution or 0)

Now we must prove that the solution exists.

\(\displaystyle 1=2(-1)^{t}+(-1)^{x}=(-1)^{x}[2(-1)^{t-x}+1]\).

\(\displaystyle \begin{cases} (-1)^{-x}=(-1)^{x}=2(-1)^{t-x}+1 \\ (-1)^{t-x}<0\equiv (-1)^{t-x}=-1 \end{cases}\).

\(\displaystyle (-1)^{x}=-1\) means that the equation \(\displaystyle x^2 -2x=(-1)^{x}\) has only one solution.

We can simply set \(\displaystyle x=(-1)^{t}\) and \(\displaystyle x-2=(-1)^{r}\) but it can be done easily below:

Since \(\displaystyle x^2 -2x = x(x-2) \) then \(\displaystyle -1\leq (x,x-2) \leq 1\) or \(\displaystyle x=(-1)^{t}\) and \(\displaystyle x-2=(-1)^{t}-2<0\).

\(\displaystyle x-2<0\) means \(\displaystyle (-1)^{t-x} <0\).

The only negative value \(\displaystyle (-1)^{t-x}\) can take is -1.(if it exists, from now there is one solution or 0)

Now we must prove that the solution exists.

\(\displaystyle 1=2(-1)^{t}+(-1)^{x}=(-1)^{x}[2(-1)^{t-x}+1]\).

\(\displaystyle \begin{cases} (-1)^{-x}=(-1)^{x}=2(-1)^{t-x}+1 \\ (-1)^{t-x}<0\equiv (-1)^{t-x}=-1 \end{cases}\).

\(\displaystyle (-1)^{x}=-1\) means that the equation \(\displaystyle x^2 -2x=(-1)^{x}\) has only one solution.

Last edited:

On the other hand, **DarnItJimImAnEngineer**'s solution is very straightforward.

$x^2-2x+1 = (x-1)^2 = 0$ has a single (repeated) solution at $x=1$ and since $(-1)^1=-1$ this is indeed a solution of the original equation.

$x^2-2x-1 = 0$ quite obviously does not have integer solutions (because $b^2-4ac = 8$ which is not a square number), so there are no solutions for even $x$.

Your statement that $x$ and $x-2$ must both be between $-1$ and $1$ is somewhat confusing - it would appear at first glance that value of non-unit magnitude for one expression would require the other to be outside that range. However, it does lead to an easier proof.

$$x^2-2x = (-1)^x \implies |x^2-2x| = |x(x-2)| = 1$$

Since $x \in \mathbb Z$, $(x-2) \in \mathbb Z$ too and thus the only possible solution is when $|x| = |x-2| = 1$ (because there are no non-zero integers with a magnitude less than 1). and this clearly happens only when $x=1$.

However, since both this method and yours require factoring the LHS of the equation, they are not, in general significantly easier than**DarnItJimImAnEngineer**'s solution. Take $x^{2}-13x=30(-1)^{x}$ as an example, or $x^2-5x+5 = (-1)^x$. The second of these doesn't easily yield to methods other than that of **DarnItJimImAnEngineer**.

$x^2-2x+1 = (x-1)^2 = 0$ has a single (repeated) solution at $x=1$ and since $(-1)^1=-1$ this is indeed a solution of the original equation.

$x^2-2x-1 = 0$ quite obviously does not have integer solutions (because $b^2-4ac = 8$ which is not a square number), so there are no solutions for even $x$.

Your statement that $x$ and $x-2$ must both be between $-1$ and $1$ is somewhat confusing - it would appear at first glance that value of non-unit magnitude for one expression would require the other to be outside that range. However, it does lead to an easier proof.

$$x^2-2x = (-1)^x \implies |x^2-2x| = |x(x-2)| = 1$$

Since $x \in \mathbb Z$, $(x-2) \in \mathbb Z$ too and thus the only possible solution is when $|x| = |x-2| = 1$ (because there are no non-zero integers with a magnitude less than 1). and this clearly happens only when $x=1$.

However, since both this method and yours require factoring the LHS of the equation, they are not, in general significantly easier than

Last edited:

Another method for this case may work .

\(\displaystyle \displaystyle x^2 -2x =(-1)^x \; , x\in \mathbb{Z}.\).

\(\displaystyle (-1)^{-x}=(-1)^{x}=\frac{1}{-x(-x-2)} \in \mathbb{Z}\).

\(\displaystyle x(x+2)\; | \; 1 \Rightarrow \begin{cases} x|1 \\ (x+2)|1 \equiv x|(-1) \end{cases} \) .

Hence \(\displaystyle x<0\) and**x** divides \(\displaystyle 1\) means there is only one solution.

\(\displaystyle \displaystyle x^2 -2x =(-1)^x \; , x\in \mathbb{Z}.\).

\(\displaystyle (-1)^{-x}=(-1)^{x}=\frac{1}{-x(-x-2)} \in \mathbb{Z}\).

\(\displaystyle x(x+2)\; | \; 1 \Rightarrow \begin{cases} x|1 \\ (x+2)|1 \equiv x|(-1) \end{cases} \) .

Hence \(\displaystyle x<0\) and

Last edited:

The method still works, but you have an extra negative sign. It should beAnother method for this case may work .

\(\displaystyle \displaystyle x^2 -2x =(-1)^x \; , x\in \mathbb{Z}.\).

\(\displaystyle (-1)^{-x}=(-1)^{x}=\frac{1}{-x(-x-2)} \in \mathbb{Z}\).

\(\displaystyle x(x+2)\; | \; 1 \Rightarrow \begin{cases} x|1 \\ (x+2)|1 \equiv x|(-1) \end{cases} \) .

Hence \(\displaystyle x<0\) andxdivides \(\displaystyle 1\) means there is only one solution.

\(\displaystyle (-1)^x = \dfrac{1}{-x( -x + 2) }\)

-Dan

\(\displaystyle (2p-1)^{2} +2(2p-1)=4p^{2}-4p+1+4p-2=(-1)^{2p-1}=-1\).

\(\displaystyle 4p^{2} -1=-1\) or \(\displaystyle p^{2}=0\) proves there is only one solution(p=0). x=2p-1=-1.

Similar Math Discussions | Math Forum | Date |
---|---|---|

Need help to verify the number of solutions | Elementary Math | |

Number of solutions | Elementary Math | |

Number of solutions | Elementary Math | |

Number of solutions | Elementary Math |