simple equation

Dec 2015
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Earth
\(\displaystyle n^2 -n =(-1)^n +(-1)^{n^2 } \: , \: n\in\mathbb{N}\).
 
Dec 2015
1,076
166
Earth
Do we have to solve it ? :D
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 : \(\displaystyle x=(-1)^x \) ?
 
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Aug 2019
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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, $n=2$.
 

skipjack

Forum Staff
Dec 2006
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For the original problem, just check $n = 1$ and $n = 2$.

For the new problem, is $x$ required to be real?
 
Dec 2015
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\(\displaystyle x=(-1)^x \implies x(-1)^x =1 \) .
\(\displaystyle (-1)^x = x^{(-1)} \implies x=-1\). :)
 
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Dec 2015
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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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Dec 2015
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Earth
(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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