Question

Find the ordered pairs of integer solutions (x,y) such that \(2^x + 1 = y^2\) .

Answer 1

@vishweshshrimali5 @Miracrown

Answer 2

Couldn't find anything easier ? O.o

Answer 3

Okay one way is to draw the curves

Answer 4

:D Well this question is quite straightforward. Just think about it! :-)

Answer 5

(3,3)

Answer 6

No doubt in that but how did you get that?

Answer 7

(3,-3)

Just trial and error method for now. :)

Answer 8

Actually not trial and error method

Answer 9

See: you have asked for integral solutions, so, 2^x + 1 must be an integer

Answer 10

That means that 2^x must also be an integer.

Answer 11

So, x must be +ve

Answer 12

Also, 2^x + 1 must be a square

Answer 13

I'm getting your point, but that is not exactly the perfect solution for this. You can think other way too.

Answer 14

May be, I will give miracrown and hartnn a chance to solve this. Till then, @vishweshshrimali5 - if you get any method, PM me. :-)

Answer 15

lol

Answer 16

you guys are brothers, why don't you do it one-on-one?

Answer 17

Actually, I have asked him this question again and again but he just prefers asking on OS !!

Answer 18

Haha.. I know whenever I will ask to my brother, he will answer it immediately, so, I just discuss with him and others regarding the questions. :-)

Answer 19

Have you given it a try @Miracrown ?

Answer 20

Well maybe he feels embarrassed asking you in a causal style.

Answer 21

May be its just that I don't like when he doesn't pay attention to me or tries to sleeps. Well it also may be because I scold him a lot while teaching him.

Answer 22

Haha! I agree. I am sleepy .. lol and you scold me a lot- which is what I like ..

OKay, we will discuss about this later, anyone knows the method ? ? ?

Answer 23

I do, but I won't help.

Answer 24

I really would have loved the graph method, but, I can't draw the graph with lappy.

Answer 25

*laptop

Answer 26

I am also not going to help you XD

Answer 27

lol @Miracrown - Why so?

Well, there is another method instead of graphing it.

Answer 28

lol ..

Okay, fine, I will put up my method here, as you guys *ARE NOT GOING TO HELP ME* :

\(2^x + 1 = y^2 \\
2^x = y^2 - 1\\
\text{Identity} \space a^2 - b^2 = (a+b)(a-b) \\
2^x = (y+1)(y-1) \)
Since the RHS has integers whose product is power of 2.

Thus, both (y+1) and (y-1) should be the powers of 2.

Now, as (y + 1 )- (y-1) = 2

therefore, it implies, y + 1 = 4 and y -1 = 2

this gives y = 3 and hence, x = 3.

(3,3) is the solution.

Answer 29

Its really late, so excuse me if I go wrong anywhere.


the problem is asking for integer solutions..

let's analyses the x-exponent, it must be an integer
let's try some values for x