Number Theory Number Theory Math Forum

 September 3rd, 2017, 02:36 AM #1 Newbie   Joined: Sep 2017 From: NY Posts: 6 Thanks: 0 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.
 September 4th, 2017, 10:19 AM #2 Member   Joined: May 2017 From: Russia Posts: 34 Thanks: 5 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. Thanks from xps32
 September 7th, 2017, 12:16 PM #3 Member   Joined: May 2017 From: Russia Posts: 34 Thanks: 5 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. Thanks from xps32
 September 10th, 2017, 01:49 PM #4 Newbie   Joined: Sep 2017 From: NY Posts: 6 Thanks: 0 Almost