APMO Problem Solution

Question

Find all positive integers $n$ such that the equation $$x^n+(2-x)^n+(2+x)^n=0$$ has integer solutions. Now for even $n$ it is straight forward that the given equation has no integer solutions. So let $n$ be odd then writing the equation as $$x^n+(x+2)^n=(x-2)^n$$ $$=>x^n=(x-2)^n-(x+2)^n$$ Let $p$ be a prime factor of $x$ Now as we can see $x-2$ is congruent to $-2 modp$ and $x+2$ is congruent to $2 modp$ So $$(x-2)^n-(x+2)^n$$ is congruent to $$(-2)^n-(2)^n modp$$ But as for all odd $n$ $$(-2)^n-(2)^n=-2^{n+1}$$ So this gives $$p|2$$ $$=>p=2$$ But as $p$ is an arbitrary prime factor of $x$ So we get $$x=p^a$$ for some natural number $a$ Now from the equation $$x^n=(x-2)^n-(x+2)^n$$ we get that $x^n=p^{an}|2^{n+1}$ so we get $$an\leq n+1$$ But as $n$ is a natural number so we get $a=2$ and $x=4$ giving $n=1$ . So for only $n=1$ the given equation has integer solutions Am I right?

Answer 1

You are totally correct up to and including the point that $p=2$ must be true.

Since $x$ can by any integer, what you can conclude then is

$$x=0, \text{ or } x=\pm 2^a; a\in \mathbb Z, a\ge 0$$

$x=0$ is a solution for no $n$, so the negative sign is the only real thing you missed.

The $p^{an}|2^{n+1}$ seems to come from nowhere, I guess it's coming from your original $x^n \equiv -2^{n+1} \pmod p$, but this is not correct.

But the idea is right, compare how many powers of $2$ divide the left and right hand side of

$$x^n=(x-2)^n-(x+2)^n$$

It turns out that the general proof works for only for $\lvert x \rvert \ge 4$, so let's assume this for a moment.

It means that

$$ \begin{eqnarray} (x-2)^n-(x+2)^n &=&2^n \left(\left(\frac{x-2}2\right)^n - \left(\frac{x+2}2\right)^n\right)\\ &=& 2^n \left(\left(\frac{x-2}2\right) - \left(\frac{x+2}2\right)\right)\sum_{i=0}^{n-1}\left(\left(\frac{x-2}2\right)^i\left(\frac{x+2}2\right)\right)^{n-1-i}\\ &=& -2^{n+1}\sum_{i=0}^{n-1}\left(\left(\frac{x-2}2\right)^i\left(\frac{x+2}2\right)\right)^{n-1-i}. \end{eqnarray} $$

Assuming $\lvert x \rvert \ge 4$ implies $4|x$, as $x$ is $\pm$a power of $2$. That means that both $\left(\frac{x-2}2\right)$ and $\left(\frac{x+2}2\right)$ are odd numbers. That means the sum on the last line is a sum of an odd number of terms ($n$) and each summand is odd, so the sum is odd as well. So you get

$$2^{n+2} | (x-2)^n-(x+2)^n \text{ is false}$$

So we know that $2^{an}$ divides $x^n$, but $2^{n+2}$ does not divided $(x-2)^n-(x+2)^n$.

This means $an \le n+1$ (as you said, but with no real justification), which is equivalent to $(a-1)n \le 1$ and since we assumed $\lvert x \rvert \ge 4$ we have $a\ge 2$ which makes this only possible if $a=2, n=1$!

We have two potential solutions $x=4, x=-4$ and it is easy to verify that $x=-4$ is a solution ($x=4$ is not, but that's not important). So for $n=1$ we actually have a solution!

Having dealt with $\lvert x \rvert \ge 4$ what remains are $x=\pm1, x=\pm2$.

$x=1$ means ($n$ is odd):

$$1=1^n=(-1)^n+3^n=3^n-1$$

which is not true for any positive $n$,as the right hand side is just bigger than the left hand side for positive $n$.

$x=-1$ means

$$-1=(-1)^n=(-3)^n+1^n=-3^n+1,$$

which doesn't work for the same reason.

So we need to consider finally only $x=\pm 2. $x=2$ means

$$2^n=0^n-4^n$$,

which is impossible for positive $n$.

$x=-2$ means (recall that $n$ is odd)

$$-2^n=(-2)^n=(-4)^n -0^n=-4^n,$$

which is equally impossible.

To sum this up, only for $n=1$ is there an integer solution in $x$, which is $x=-4$.

Answer 2

Note that $(2-x)$ and $(2+x)$ have the same parity. Then $(2-x)^n+(2+x)^n=-x^n$ is even and hence $x=2k$ for some $k$.

The equation then becomes $$k^n+(1-k)^n+(1+k)^n=0$$

As you said $n$ is odd and expanding you get $$2+2\binom{n}{2}k^2+2\binom{n}{4}k^4+..+2\binom{n}{n-1}k^{n-1}+k^n=0$$

Now, $k^n$ is even, thus $k=2l$ is even. Then $$2\binom{n}{2}k^2+2\binom{n}{4}k^4+..+2\binom{n}{n-1}k^{n-1}=-2-k^n$$

is a multiple of $4$, showing that $4|2+(2l)^n$. It is easy to conclude that $n=1$.