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March 23rd, 2019, 06:22 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81  A type of equation
Find all integers n such that $\displaystyle n^2 +n$ is a power of $\displaystyle 2$ .

March 23rd, 2019, 08:07 AM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,194 Thanks: 897 Math Focus: Wibbly wobbly timeywimey stuff.  
March 23rd, 2019, 11:25 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81 
Let $\displaystyle n=2^x $ This means $\displaystyle 1+2^x $ must be an even number . Which happens only when $\displaystyle 2^x =1$ or $\displaystyle x=0$ . $\displaystyle n=2^x =1$ . But this is an easy specific case to solve it like the way above . Last edited by idontknow; March 23rd, 2019 at 11:43 AM. 
March 23rd, 2019, 01:00 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337  Quote:
The only two adjacent integers that are both powers of 2 are $2^0=1,~2^1=2$ $n=0$  
March 23rd, 2019, 02:21 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,194 Thanks: 897 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
March 25th, 2019, 06:09 AM  #6 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81 
The other method is by applying the number of solutions which is 1 . $\displaystyle 2^x \cdot 2^y =2^k \; $ , $\displaystyle x+y=k$ . The number of solutions is $\displaystyle N(k)=k+1 \lfloor \frac{k+1}{2} \rfloor =1$ or $\displaystyle k=\lfloor \frac{k+1}{2} \rfloor $ , $\displaystyle k>0$ . By the integer part inequality $\displaystyle k\leq 2k+1 \leq k+2$ $\displaystyle \Rightarrow \; k=1$ or $\displaystyle x+y=1$ . By substitution to the first equation $\displaystyle n^2 +n =2$ (quadratic equation) which has solution $\displaystyle n=1$ . Last edited by idontknow; March 25th, 2019 at 06:14 AM. 
March 25th, 2019, 06:32 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
Think again  there are two solutions.

March 25th, 2019, 06:37 AM  #8 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81 
I mean there is only one pair $(x,y)$ that satisfies $x+y=k$. Edit: the other solution is $n=2$, but I solved for positive integers. Last edited by skipjack; March 25th, 2019 at 10:12 AM. 
March 25th, 2019, 10:11 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
The original post didn't require $n$ to be positive. I assume that romsek intended $x = 0$ or $n = 1$. 
March 25th, 2019, 12:20 PM  #10 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81 
I just posted my approach which doesn't always give solutions. Here is the way it should be solved: ninteger, kpositive integer including 0. $\displaystyle n^2 +n 2^k =0$ (quadratic equation ). $\displaystyle n=\frac{1 \pm \sqrt{D} }{2}$, where $\displaystyle D=1+2^{2+k}$. For which values of k is the discriminant the square of an integer? It can be seen that $\displaystyle k=1$ is a value that makes the discriminant the square of an integer. How to find k with method? Last edited by skipjack; March 25th, 2019 at 09:33 PM. 

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