# simple equation

#### idontknow

$$\displaystyle n^2 -n =(-1)^n +(-1)^{n^2 } \: , \: n\in\mathbb{N}$$.

$$\displaystyle n^2 -n =(-1)^n +(-1)^{n^2 } \: , \: n\in\mathbb{N}$$.
Do we have to solve it ? • topsquark

#### idontknow

Do we have to solve it ? let me solve it .
$$\displaystyle n^2 - n =n(n-1)= 2(-1)^n \implies n-1=1$$.
since $$\displaystyle n(n-1) \geq 0$$ and 2 is prime , equal n-1 with 1 because n , n-1 are adjacent integers so n=2 . btw how would you solve : $$\displaystyle x=(-1)^x$$ ?

• Okay, I will take quadratic equation approach. Let’s first assume that $n$ is even $$n^2 - n = (-1)^n + (-1)^{n^2} \\ n^2 -n = 1 + 1 \\ n^2 - n -2 = 0 \\ \text{The roots of the above equation are : 2 and -1. Since, n can only be a natural number we would get n=2}$$

If, however, $n$ is odd, then $$n^2 -n = -1 -1 \\ n^2 -n +2 =0 \\ \text{This equation doesn’t have any real roots}$$
Hence, $n=2$.

• topsquark and idontknow

btw how would you solve : $$\displaystyle x=(-1)^x$$ ?
Is $x$ a natural number, rational or real?

#### skipjack

Forum Staff
For the original problem, just check $n = 1$ and $n = 2$.

For the new problem, is $x$ required to be real?

x-integer .

#### idontknow

$$\displaystyle x=(-1)^x \implies x(-1)^x =1$$ .
$$\displaystyle (-1)^x = x^{(-1)} \implies x=-1$$. • topsquark

#### idontknow

I was asking for method to solve the equation by not assuming or checking whether -n- is odd or even .
Another equation with 2-variables :
(1) $$\displaystyle 2mn = (-1)^{m+n}+n \; \; ; m ,n \in \mathbb{N}.$$

(2) How many real solutions the equation has ? $$\displaystyle \; \sin x = x/7 .$$ (answer is 3 )

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#### idontknow

(2) for sinx=x/n , n-natural , then $N \eqsim 4 \left\lceil \frac{n-1}{2\pi} \right\rceil - 1$ , n=7 .
$$\displaystyle N\eqsim 4 \left\lceil \frac{3}{\pi} \right\rceil-1=3$$.
maybe we can get a better approximation by $\sin(x) = x\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2\pi^2}\right) =x/n$ , x=0 cancels so we have 1 solution.
$\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2\pi^2}\right)-1/7 =0$ remains to give 2 solutions by applying derivative of $\dfrac{\sin(\pi x)}{\pi x} = \prod_{k=1}^\infty \left(1-\frac{x^2}{k^2}\right)$ etc.

(1) LHS-even implies n-odd implies m-odd . $$\displaystyle 2mn=1+n \implies n\leq 1$$ or n=1 , $$\displaystyle \; \; 2m=2$$ or m=n=1.
also searching solution for m=n finds 1 as solution.

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