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September 3rd, 2017, 02:36 AM   #1
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Hey!
\begin{align*}
(x+1)^{n+1}-(x-1)^{n+1}&=y^n
\end{align*}
There is no solutions in integers x, y, n.
n>1
How to show it?
You have a hint?

Last edited by skipjack; September 3rd, 2017 at 02:10 PM.
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September 4th, 2017, 10:19 AM   #2
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A particular case when $\displaystyle x$ is odd.

Suppose that $\displaystyle x, y\in\mathbb{Z}$ are the solution of equation, and $\displaystyle x$ is odd.

Let $\displaystyle x=t+1$. So, $\displaystyle t$ is even.

$\displaystyle (t+2)^{n+1}-t^{n+1} = y^n.$

One see that y is even too. Divide the equation with $\displaystyle 2^{n+1}$:

$\displaystyle \left(\frac{t}{2}+1\right)^{n+1}-\left(\frac{t}{2}\right)^{n+1} = \frac{ \left(\frac{y}{2}\right)^{n}}{2}$.

On the left side one has an integer. It means that $\displaystyle \frac{y}{2}$ is even and, recalling that $\displaystyle n\gt 1$, $\displaystyle \frac{\left(\frac{y}{2}\right)^{n}}{2}$ is even too.

One has a contradiction: there is an odd integer on the left side of the equation and an even integer on the right side.
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September 7th, 2017, 12:16 PM   #3
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And another particular case when $\displaystyle x$ is even but $\displaystyle n$ is even too.

Suppose that $\displaystyle x,y\in\mathbb{Z}$ are the solution of the equation and x is even.
Let $\displaystyle x=t+1$. So, $\displaystyle t$ is odd.
$\displaystyle (t+2)^{n+1}-t^{n+1} = y^n.$
In this case $\displaystyle y$ is even too.
Using binomial theorem, one obtains
$\displaystyle \left( \sum_{k=0}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} t^{n+1-k} \cdot 2^k \right) - t^{n+1} =y^n.$
$\displaystyle \sum_{k=1}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} t^{n+1-k} \cdot 2^k =y^n.$
$\displaystyle (n+1)\cdot 2\cdot t^{n} =y^n - \sum_{k=2}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} t^{n+1-k} \cdot 2^k.$
The integer on the left side of the equality is divisible by 2 but not divisible by 4. On the right side one has an integer divisible by 4. It's a contradiction.
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September 10th, 2017, 01:49 PM   #4
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