1. Sep 14, 2008

### özgürden

Determine all pairs (x, y) of integers such that
1 + 2^x + 2^(2x+1) = y^2.

^=Exponent

2. Sep 14, 2008

### chaoseverlasting

If you assume $$2^x=z$$ then the equation reduces to $$1+z+2z^2=y^2$$.
Completing the square results in $$(4z+1)^2+7=8y^2$$. Substituting $$z=2^x$$ and taking log2 gives,

$$x+2=log2(-1 +-\sqrt{8y^2-7})$$.

For all integral values of x, LHS is an integer , say k.

Now, this gives,

$$2^k=-1+-\sqrt{8y^2-7}$$ which is also an integer (say k1) because 2k is also an integer. Which means that $$\sqrt{8y^2-7}$$ is also an integer (say k2).

So thats the final condition you arrive at. For all integer values of y for which $$\sqrt(8y^2-7)$$ is an integer, x and y satisfy the above condition.

3. Sep 14, 2008

### chaoseverlasting

Dont know how to get the latex code right, but -1+-\sqrt{8y^2-7} means the positive and negative root of 8y^2-7 subtracted by 1.

4. Sep 14, 2008

### chaoseverlasting

By hand, y= 1,2,4 seems to satisfy the above condition, but analytically, maybe someone else could come up with a better solution. Dont know how to factor in the integer square condition.

5. Sep 14, 2008

### dodo

Here is an alternate (unfinished) route.

A value of x = -1 will produce a LHS of 2, which is not a square; negative integers < -1 will not produce an integer LHS. Thus we need only be concerned with integers x >= 0.

Trying out a few values by hand, we see that the pairs (0,2) and (4,23) are solutions, and x=1,2,3 are not. So now the question becomes, is there another solution for x > 4?

With a little algebra (subtract 1 and expand the RHS) the equation becomes
$$2^x (2^{x+1} + 1) = (y-1)(y+1)$$​
thus the RHS is the product of two consecutive evens, and y is odd. Let y = 2k+1; substituting, the RHS will become 2k(2k+1) = 4k(k+1). Dividing by 4 and replacing n = x-2 (with n > 2), we get
$$2^n (2^{n+3} + 1) = k(k+1)$$​
Now one of k or k+1 is odd; the other will be divisible by 2^n (as a matter of fact, since 2^{n+3} + 1 is odd, 2^n is the greatest power of 2 that divides the RHS).

If we could prove that no integer n > 2 satisfies the above, we would be done. The requirement of exactly n 2's among the prime factors of the RHS only sieves candidates, but does not provide a bound for k.

My hunch is that something has to become broken when the powers of two get bigger, i.e. that the LHS cannot be expressed as a product of two consecutive integers (for n=2 it can, though).

6. Sep 15, 2008

### snipez90

Ok, I will attempt to finish Dodo's solution. Just to recap, he subtracted 1 from both sides of the equation to factor the difference of squares. Since the LHS is even, it's not hard to see that y must be odd so he set y = 2k + 1 to arrive at

$$2^x(2^{x+1} + 1) = 2k(2k+2) \Rightarrow 2^{x-2}(2^{x+1}+1) = k(k+1)$$.

Furthermore, he set $$n = x - 2$$ but this time I'm going to put a different condition on n so that $$n \geq -2$$ because we want to find solutions for $$x \geq 0$$.

So after applying the conditions, we have

$$2^n(2^{n+3} + 1) = k(k+1), n \geq -2$$

Now consecutive integers are relatively prime and therefore, $$2^n | k$$ or $$2^n | k+1$$.

Case 1: $$2^n | k \Rightarrow k = a2^n$$ for some integer $$a > 0$$. Plugging this back into our last equation, we get $$2^n(2^{n+3}+1) = a2^n(a2^n + 1) \Rightarrow a^22^n + a = 2^{n+3} + 1 \Rightarrow 2^n(a^2 - 2^3) = 1 - a \Rightarrow 2^n = \frac{1-a}{a^2 - 8}$$. For $$a > 2$$, the RHS is negative, which is a contradiction. $$a$$ cannot be 1 either, but for $$a = 2$$, we have $$2^n = \frac{1}{4} \Rightarrow n = -2 \Rightarrow x = 0$$. Hence (0,2) and (0, -2) are solutions.

Case 2: $$2^n | k + 1 \Rightarrow k = b2^n - 1$$ for some integer $$b > 0$$. Applying the same technique as in the first case, we have $$2^n(2^{n+3} + 1) = (b2^n - 1)(b2^n) \Rightarrow b^22^n - b = 2^{n+3} + 1 \Rightarrow 2^n(b^2 - 8) = 1 + b \Rightarrow 2^n = \frac{1+b}{b^2 - 8}$$. For $$b > 3$$, it's easy to see we won't find any solutions for n. Similarly, we won't find any solutions for $$b < 3$$ since the RHS will be negative (except our trivial solution which occurs when $$b = -2$$, but we assumed allowably $$b > 0$$). Thus, we find that $$b = 3 \Rightarrow n = 2 \Rightarrow x = 4$$ gives us our other pair of solutions, (4, 23) and (4, -23)

Note that we find our solutions in pairs because of the $$y^2$$ term in the original equation so (x, -y) is a solution whenever (x, y) is. Thus our solutions are (0, 2), (0, -2), (4, 23), and (4, -23)

7. Sep 15, 2008

### robert Ihnot

There is a small mistake in Dodo, where he says: the RHS will become 2k(2k+1) = 4k(k+1), since he really means 2k(2k+2) = 4k(k+1).

I dont understand this use of RHS, or is it LHS? Oh! From reading snipez90, I see it means nothing more than left or right side of the equation.

Last edited: Sep 15, 2008
8. Sep 15, 2008

### özgürden

CORRECT ANSWER :(x, y)- (0, 2), (0,−2), (4, 23), (4,−23)

9. Sep 15, 2008

### chaoseverlasting

Thats it? Just 4 solutions? Shouldnt there be a general formula or something?

10. Sep 17, 2008

### robert Ihnot

chaoseverlasting: Thats it? Just 4 solutions? Shouldnt there be a general formula or something?

Snipez90 did good work on that problem, I don't find any errors. So I guess we just have to accept it.