IIT JEE , AIEEE Forum - Mathematics , Algebra

shajikoshy

Find all the values of (x,y) for the equation.

1+ 2^x + 2^{2x +1} = y²

konichiwa2x

...

is always greater than 0 for all values of 'x'.. Hence, for every value of x belonging to real numbers, you will have 2 corresponding values of y. There are infinite number of solutions...

shajikoshy

IF (x,y) are integers . How to solve the equation ?

konichiwa2x

edited : check my last post down below for solution

shajikoshy

your method looks OK but
you have got the eqation wrong this time.

sprinkle

shajikoshy!

first of all 2^x is defined and positive for all x => x

R

Now, let 2^x = t

=> 1 + 2^x + 2^(2x+1) = 1 + 2^x + 2 * 2^(2x) = 1 + 2^x + 2 * (2^x)^2 = 1 + t + 2t^2

let f(t) = 2t^2 + t + 1

it has minima at t = -1/4, f(-1/4) = 7/8 and always increasing after that. (actually an opening upward parabola with vertex at (-1/4, 7/8)

but t = 2^x which can't be -ve, so we will consider minimum possible value of t

that is when x-> -

=> 2^x -> 0 => t -> 0

f(t) -> 1 when t -> 0

so f(t) has range (1,

) => y^2 = f(t) has same range => y has same range

Thus,

x

R and y

(1,

) ANS

sorry for any miscalculations!!!

konichiwa2x!

why are you taking 1 + 2^x + 2^(x+1), it is 1 + 2^x + 2^(2x+1)

shajikoshy

I appreciate your efforts to solve this problem. The answer i was looking for (x,y) was
(0,2), (0,-2), (4, 23), (4,-23) which are the only solution in integers.
I found this by trial and error. But i still believe their is a way to solve this equation.

konichiwa2x

yes sorry...made a mistake...i'll try to do it another way and get back to u (got an idea..but let me try it out now)

konichiwa2x

ok I think i got it!

Clearly (x,y) = (0,±2) gives one pair of solutions, and there are no solutions for x=1. Equally clearly there can be no solutions with x<0.

If x>1, let

. Then

,

or

.

Therefore

. So one of the numbers y+1+w, y-1-w must be a power of 2, and the other must be 7 times a power of 2.

Suppose that

. Then

. Thus

. The left side of this equation is an odd multiple of 2, hence so is the right side, and this can only happen if n=1. Then

, from which m=5 and x=4. This gives the solutions (x,y)=(4,±23).

The other possibility is that

. A similar analysis to the previous paragraph leads to the equation

. But a power of 2 can only be congruent to 1, 2 or 4 (mod 7), not to -1. So this equation has no integer solutions.

Therefore the only solutions are (x,y) = (0,±2) or (4,±23)!

konichiwa2x

and i edited my prev post.. sorry for the mistake in my first attempt.

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