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March 23rd, 2019, 06:22 AM   #1
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A type of equation

Find all integers n such that $\displaystyle n^2 +n$ is a power of $\displaystyle 2$ .
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March 23rd, 2019, 08:07 AM   #2
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Originally Posted by idontknow View Post
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?

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March 23rd, 2019, 11:25 AM   #3
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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.
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March 23rd, 2019, 01:00 PM   #4
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Quote:
Originally Posted by idontknow View Post
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

$2^0=1,~2^1=2$

$n=0$
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March 23rd, 2019, 02:21 PM   #5
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Originally Posted by idontknow View Post
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.

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March 25th, 2019, 06:09 AM   #6
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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.
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March 25th, 2019, 06:32 AM   #7
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Think again - there are two solutions.
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March 25th, 2019, 06:37 AM   #8
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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.
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March 25th, 2019, 10:11 AM   #9
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The original post didn't require $n$ to be positive.

I assume that romsek intended $x = 0$ or $n = 1$.
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March 25th, 2019, 12:20 PM   #10
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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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