# 3x2^m + 1 = n^2 - find all values such that this is true

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• Jul 2nd 2011, 09:06 PM
pikachu26134
3x2^m + 1 = n^2 - find all values such that this is true
I'm having difficulty remembering wot to do with an equation lyk this, im pretty sure its slightly different to the last one i posted and i just cant remember or find it in my textbooks. Any help would b fantastic.

3 x (2^x) + 1 = n^2

You must find all pairs of positive integers that work for this equation.
Cant remember wot method 2 use.
• Jul 2nd 2011, 11:09 PM
Also sprach Zarathustra
Re: 3x2^m + 1 = n^2 - find all values such that this is true
Quote:

Originally Posted by pikachu26134
I'm having difficulty remembering wot to do with an equation lyk this, im pretty sure its slightly different to the last one i posted and i just cant remember or find it in my textbooks. Any help would b fantastic.

3 x (2^x) + 1 = n^2

You must find all pairs of positive integers that work for this equation.
Cant remember wot method 2 use.

$3\cdot 2^x =n^2-1$

$3\cdot 2^x =(n-1)(n+1)$

So, what could you say?
• Jul 2nd 2011, 11:15 PM
pikachu26134
Re: 3x2^m + 1 = n^2 - find all values such that this is true
Sorry, i dont 100% know wot to do from here. Im a bit slow at this sort of thing.
• Jul 3rd 2011, 12:38 AM
Also sprach Zarathustra
Re: 3x2^m + 1 = n^2 - find all values such that this is true
Quote:

Originally Posted by pikachu26134
So, what do i do from there may i ask?

x even positive integer and $n=2^x-1$

or:

x odd positive integer and $n=2^x+1$
• Jul 3rd 2011, 12:43 AM
pikachu26134
Re: 3x2^m + 1 = n^2 - find all values such that this is true
Still not getting u - how do u get the actual numerical values??
• Jul 3rd 2011, 12:50 AM
Also sprach Zarathustra
Re: 3x2^m + 1 = n^2 - find all values such that this is true
Quote:

Originally Posted by pikachu26134
Still not getting u - how do u get the actual numerical values??

n-1 or n+1 must be equal to 2^x , why?

if n-1 = 2^x ==> n= 2^x +1 BUT you gave also a 3 to deal with...
• Jul 3rd 2011, 12:51 AM
pikachu26134
Thats wot im having problems with. Is it lyk an infinite number of pairs??

so would i attain 3 x (2^m) + 1 = n^2
Would i substitute n's value with terms of m, or would i just say theres an infinite number of pairs?

Im still working on it, cant think wot it is, ideas??
• Jul 3rd 2011, 02:31 AM
Also sprach Zarathustra
Re: 3x2^m + 1 = n^2 - find all values such that this is true
Quote:

Originally Posted by pikachu26134
Im still working on it, cant think wot it is, ideas??

$3\cdot 2^x+1=n^2$

$3\cdot 2^x=n^2-1$

$3\cdot 2^x=(n-1)(n+1)$

First case:

$n-1=2^x$

so, $n=2^x+1$

But on the left side we have also $3$, hence $n+1$ must have a factor of $3$, but when this happens? It happens when $n+1=2^x+2$ have a factor of $3$ :

$2^x+2=2(2^{x-1}+1)$

hence, $2^{x-1}+1$ have a factor of $3$, but for which $x$?

It easy to show that $x$ must be odd.

Now, do the same for the second case, but now when $n+1=2^x$.

At the and you will get a collection of infinite solutions.
• Jul 3rd 2011, 02:57 AM
Isomorphism
Re: 3x2^m + 1 = n^2 - find all values such that this is true
Quote:

Originally Posted by Also sprach Zarathustra
n-1 or n+1 must be equal to 2^x , why?

if n-1 = 2^x ==> n= 2^x +1 BUT you gave also a 3 to deal with...

That is not true. n=5 , x=3 shows $5^2 - 1 = 24 = 3. 2^3$. but neither n-1, nor n+1 is 8.

pikachu26134,
Here are some exercises for you,
1) Show that if n > 2 satisfies your equation,n must be odd.
1') Substitute n=2 and see if it satisfies your equation.
2) Show that $(n-1)(n+1) = 3.2^x$ if and only if $n-1 = 3.2^k , n + 1 = 2^{x-k}$ OR $n-1 = 2^k , n + 1 = 3.2^{x-k}$ for some integer k such that $0 \leq k \leq x$
3) By subtracting the equations in the above system individually, conclude that k=1 is the only possible scenario. Thus arrive at the conclusion that only two solutions with odd 'n' are possible.

So totally there are 3 solutions. Show me some effort by trying to solve the above simpler exercises and I will help you further.