September 3rd, 2017, 03:36 AM  #1 
Newbie Joined: Sep 2017 From: NY Posts: 6 Thanks: 0 
Hey! \begin{align*} (x+1)^{n+1}(x1)^{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 03:10 PM. 
September 4th, 2017, 11:19 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. 
September 7th, 2017, 01:16 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 $\displaystyle \left( \sum_{k=0}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} t^{n+1k} \cdot 2^k \right)  t^{n+1} =y^n.$ $\displaystyle \sum_{k=1}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} t^{n+1k} \cdot 2^k =y^n.$ $\displaystyle (n+1)\cdot 2\cdot t^{n} =y^n  \sum_{k=2}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} t^{n+1k} \cdot 2^k.$ 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. 
September 10th, 2017, 02:49 PM  #4 
Newbie Joined: Sep 2017 From: NY Posts: 6 Thanks: 0 
Almost 

Tags 
deadlock 
Thread Tools  
Display Modes  
