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

1. ## 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.

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

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?

3. ## 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.

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

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 $n=2^x+1$

5. ## 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??

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

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...

7. 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??

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

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.

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

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.