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 March 23rd, 2019, 06:22 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 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
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Quote:
 Originally Posted by idontknow Find all integers n such that $\displaystyle n^2 +n$ is a power of $\displaystyle 2$ .
If $\displaystyle n^2 + n = n(n + 1)$ is a power or 2 then both n and n + 1 have to be powers of 2. When does this happen?

-Dan

 March 23rd, 2019, 11:25 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 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
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Quote:
 Originally Posted by idontknow 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 .
TopSquark told you how to solve it.

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
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Quote:
 Originally Posted by idontknow 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 .
Honestly I think it's a ridiculous question. (It is not, however, "Elementary Math"). It's designed to make you think that it is much harder than it actually is and I've never found riddles to have much educational value.

-Dan

 March 25th, 2019, 06:09 AM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 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: 20,968 Thanks: 2217 Think again - there are two solutions. Thanks from idontknow
 March 25th, 2019, 06:37 AM #8 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 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: 20,968 Thanks: 2217 The original post didn't require $n$ to be positive. I assume that romsek intended $x = 0$ or $n = 1$. Thanks from topsquark and idontknow
 March 25th, 2019, 12:20 PM #10 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 I just posted my approach which doesn't always give solutions. Here is the way it should be solved: n-integer, k-positive 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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