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.

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

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**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.

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

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

So, what could you say?

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.

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

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**pikachu26134** So, what do i do from there may i ask?

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

or:

x odd positive integer and $\displaystyle n=2^x+1$

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??

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

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**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...

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

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**pikachu26134** Im still working on it, cant think wot it is, ideas??

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

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

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

First case:

$\displaystyle n-1=2^x$

so, $\displaystyle n=2^x+1$

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

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

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

It easy to show that $\displaystyle x$ must be **odd**.

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

At the and you will get a collection of infinite solutions.

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

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**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 $\displaystyle 5^2 - 1 = 24 = 3. 2^3$. 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 $\displaystyle (n-1)(n+1) = 3.2^x$ if and only if $\displaystyle n-1 = 3.2^k , n + 1 = 2^{x-k}$ OR $\displaystyle n-1 = 2^k , n + 1 = 3.2^{x-k}$ for some integer k such that $\displaystyle 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.