Determine all positive integers $n$ for which the equation \[ x^n + (2+x)^n + (2-x)^n = 0 \] has an integer as a solution.
Problem
Source: APMO 1993
Tags: number theory unsolved, number theory
Megus
11.03.2006 19:02
As $n$ even clearly doesn't work we can suppose that $n=2k+1$. As $k=0$ clearly works, we might suppose that $k \geq 1$. Now: $x^{2k+1}+4((2+x)^{2k}-....+(2-x)^{2k})=0$ so $2|x$ let $x \to 2x$ and we have: $2^{2k-1}x^{2k+1}+2^{2k}((1+x)^{2k}-...+(1-x)^{2k})=0$ so again $2|x$ and we let $x \to 2x$. Finally: $2^{2k}x^{2k+1}+(1+2x)^{2k}-(1-4x^2)(1+2x)^{2k-2}+....+(1-2x)^{2k} \equiv (1+2x)^{2k}-1+(1-2x)^{2k}\equiv 1 \not \equiv 0 \mod 4$ Contradiction. So we only $n=1$ satisfies given condition.
Davron
12.03.2006 06:26
Megus please could you explain your solution... Davron
Megus
12.03.2006 14:18
what do you feel uncertain about?
towersfreak2006
14.06.2006 00:24
shobber wrote: Determine all positive integers $n$ for which the equation \[ x^n + (2+x)^n + (2-x)^n = 0 \] has an integer as a solution.
Suppose $n$ is even. Then \[ x^n+(2+x)^n+(2-x)^n\ge0, \] with equality at $x=2+x=2-x=0$, which clearly, can not be obtained. Hence, $n$ cannot be even.
Suppose $n$ is odd. Then \[ x^n + (2+x)^n + (2-x)^n = 0 \] implies \[ x^n+(2+x)^n=(x-2)^n. \] Consider $n=1$. Then \[ x+(2+x)=(x-2) \] implies $x=-4$.
Now suppose $n\ge3$. If $x>2$, then from Fermat's Last Theorem, there are no integer solutions. The same conclusion follows if $x<-2$ and both sides are multiplied by $-1$. Furthermore, $x=2$, $x=1$, $x=0$, $x=-1$, and $x=-2$ have no solutions, upon inspection.
Therefore, the only valid $n$ for which the equation has integer solutions in $x$ is $\boxed{n=1}$.
persya
22.07.2010 08:31
I think megus's solution is very easy to understand because his solution is similar to the proof of Fermat to show that the equation $ x^4+y^4=z^2 $ has no integer roots
mathmdmb
22.07.2010 11:09
shobber wrote: Determine all positive integers $n$ for which the equation \[ x^n + (2+x)^n + (2-x)^n = 0 \] has an integer as a solution. Trivial.Since $n$ positive integer,we see two cases.First $n$ even and the result is obvious $x=0$.now see if $n$ even,we can use fermat's last theorem
peregrinefalcon88
26.01.2011 23:33
We can see that n is odd. We can also notice that x is negative so let y = -x. We have $-y^n+(2-y)^n+(2+y)^n = 0 \rightarrow (y+2)^n = y^n+(y-2)^n$.(apply fermats last theorem) Considering this equation mod y we see that $y = 2^k$. Letting 2z = y we obtain the new equation $(z+1)^n = z^n + (z-1)^n$ and considering this mod z yields z = 1, 2. y = 2 means n = 0. y = 4 means n = 1.
Amir Hossein
26.02.2011 19:09
Amazing, this problem is APMO 1993 (this topic) and IMO LongList 1988 Problem 28.
CakeIsEaten
04.03.2011 09:36
why does this problem remind me of fermat's last...?
anonymouslonely
03.08.2011 12:41
if $ n $ is even then $ x=2-x=2+x=0 $ false. if $ n $ is odd then $ (2+x)^{n}+(2-x)^{n} $ is even, $ 0 $ is even so $ x $ is even. take $ x=2w $ so $ w^{n}+(1+w)^{n}+(1-w)^{n}=0 $ but $ (1+w)^{n} $ is congruent with 1 mod $ w $ and $ (1+w)^{n} $ is congruent with 1 mod $ w $ and $ 0 $,$ w^{n} $ are divisible by $ w $ so $ 2 $ is divisible by $ w $. $ w $ is even, the proof of this is similar like the proof for $ x $. 1. $ w=2 $ so $ 2^{n}+3^{n}=1 $ impossible. 2. $ w=-2 $ so $ 2^{n}+1=3^{n} $ but $ 2^{n}+1<3^{n} $ for $ n>1 $ so $ n=1 $.