There are various ways to do it. This may be the most 'intuitive'.

I used 2^(2x+1)+2^x+1=y^2 for my working; it makes no difference. Nor does nonnegative integer(s) make much difference since negative values of x need not detain us and y can be either positive or negative.

Starting  with the segment 2^(2x+1)+2^x, write this as 2^x(2^(x+1)+1). Clearly, the ratio of 2^x to (2^(x+1)+1) is very close to 1:2, and gets ever closer to, but can never exactly be 1:2, as x increases.

Now we have 2^x(2^(x+1)+1)=(y+1)(y-1). As the two factors on LHS are so close, let 2^x=a, so that we have (approximately) a(2a)=(y+1)(y-1), or (approximately)2a^2=y^2-1, which looks familiar, and has known solutions. It follows at once that as 2^x(2^(x+1)+1) approximates 2a^2, it is not possible for a corresponding y^2 to be integral, because that could only be so in the event that (2^(x+1)+1)=2*a, which we know to be false, since they differ by 1.

So if there are solutions to 2^x(2^(x+1)+1)=(y+1)(y-1), they are (1) small, and (2) sporadic. Actually, after only 14 terms, ((1/2)*(2^(2x+1)+2^x+1)^(1/2) settles almost exactly to (integer+0.25),* so 'small' is smaller than x=14.        
        
An excel check now shows that the only solutions are {x,y}={0, 2 or -2},{4,23 or -23}. It is possible to set t=2^x, then use 2t^2+t+1-y^2=0, a solveable Pellian where we seek t's that are powers of 2, but it is probably quicker to just perform the excel check**.

*Hence the unlikely-looking ((1/2)*(2^(2x+1)+2^x+1))^(1/2)-1/4 is almost exactly an integer power of 2, and it follows that (approximately)  2^x-((1/2)*(2^(2x+1)+2^x)+1)^(1/2)+1/4 = 0), suggesting a possible alternative line of approach.

** But it does also follow that the sole solutions for 2t^8+t^4+1=y^2 are: t={0,1 or -1,2 or -2}

Edited on September 15, 2012, 5:23 am

Posted by broll on 2012-09-15 03:06:34