APL (7)

Yes, that's 7 bytes. Assume for the moment that I'm using IBM codepage 907 instead of Unicode and then each character is a byte :)

0=1|2⍟⎕

i.e. 0 = mod(log(input(),2),1)

GolfScript, 11 (for 1 (true) and 0 (false))

.,{2\?}%?0>

Put the number on the stack and then run.

GolfScript, 22 (for Yes/No)

.,{2\?}%?0>'Yes''No'if

I love how converting 1/0 to Yes/No takes as much code as the challenge itself :D

Warning: EXTREMELY inefficient ;) It does work fine for numbers up to 10000, but once you get that high you start to notice slight lag.

Explanation:

• .,: turns n into n 0..n (. duplicate, , 0..n range)
• {2\?}: to the power of 2
• %: map "power of 2" over "0..n" so it becomes n [1 2 4 8 16 ...]
• ?0>: checks to see if the array contains the number (0 is greater than index)

GolfScript, 6 chars, no decrements

~.3/&!

Here's a solution that doesn't use the x & (x-1) method in any form. It uses x & (x/3) instead. ;-) Outputs 0 if false, 1 if true.

Explanation:

• ~ evals the input string to turn it into a number,
• . duplicates it (for the subsequent &),
• 3/ divides it by three (truncating down),
• & computes the bitwise AND of the divided value with the original, which will be zero if and only if the input is zero or a power of two (i.e. has at most one bit set), and
• ! logically negates this, mapping zero to one and all other values to zero.

Notes:

• Per the clarified rules, zero is not a valid input, so this code is OK, even though it outputs 1 if the input is zero.

• If the GolfScript decrement operator ( is allowed, then the 5-character solution ~.(&! posted by aditsu is enough. However, it seems to go against the spirit of the rules, if not the letter.

• I came up with the x & (x/3) trick years ago on the Fun With Perl mailing list. (I'm sure I'm not the first to discover it, but I did (re)invent it independently.) Here's a link to the original post, including a proof that it actually works.

Mathematica 28

Numerator[Input[]~Log~2]==1

For integer powers of 2, the numerator of the base 2 log will be 1 (meaning that the log is a unit fraction).

Here we modify the function slightly to display the presumed input. We use # in place of Input[] and add & to define a pure function. It returns the same answer that would be returned if the user input the number in the above function.

    Numerator[#~Log~2] == 1 &[1024]
    Numerator[#~Log~2] == 1 &[17]

True
False

Testing several numbers at a time.

    Numerator[#~Log~2] == 1 &&/@{64,8,7,0}

{True, True, False, False}

GolfScript, 5

Outputs 1 for true, 0 for false. Based on user3142747's idea:

~.(&!

Note: ( is decrement, hopefully it doesn't count as - :)
If it does (and the OP's comments suggest that it might), then please refer to Ilmari Karonen's solution instead.
For Y/N output, append 'NY'1/= at the end (7 more bytes).

Octave (15 23)

EDIT: Updated due to user input requirement;

~mod(log2(input('')),1)

Lets the user input a value and outputs 1 for true, 0 for false.

Tested in Octave, should work in Matlab also.

R, 13 11

Based on the Perl solution. Returns FALSE or TRUE.

!log2(i)%%1

The parameter i represents the input variable.

An alternative version with user input:

!log2(scan())%%1

Perl 6 (17 characters)

say get.log(2)%%1

This program gets a line from STDIN get function, calculates logarithm with base 2 on it (log(2)), and checks if the result divides by 1 (%%1, where %% is divides by operator). Not as short as GolfScript solution, but I find this acceptable (GolfScript wins everything anyway), but way faster (even considering that Perl 6 is slow right now).

~ $ perl6 -e 'say get.log(2)%%1'
256
True
~ $ perl6 -e 'say get.log(2)%%1'
255
False

JavaScript, 41 40 characters

l=Math.log;alert(l(prompt())/l(2)%1==0);

How this works: you take the logarithm at base 2 using l(prompt()) / l(2), and if that result modulo 1 is equal to zero, then it is a power of 2.

For example: after taking the logarithm of 8 on base 2, you get 3. 3 modulo 1 is equal to 0, so this returns true.

After taking the logarithm of 7 on base 2, you get 2.807354922057604. 2.807354922057604 modulo 1 is equal to 0.807354922057604, so this returns false.

Javascript (37)

a=prompt();while(a>1)a/=2;alert(a==1)

Simple script that just divides by 2 repeatedly and checks the remainder.

Mathematica (21)

IntegerQ@Log2@Input[]

Without input it is a bit shorter

IntegerQ@Log2[8]

True

IntegerQ@Log2[7]

False

I decided to to use another approach, based on the population count or sideways sum of the number (the number of 1-bits). The idea is that all powers of two have exactly one 1 bit, and no other number does. I added a JavaScript version because I found it hilarious, though it certainly won't win any golfing competition.

J, 14 chars (outputs 0 or 1)

1=+/#:".1!:1]1

JavaScript, 76 chars (outputs true or false)

alert((~~prompt()).toString(2).split("").map(Number).filter(Boolean).length)

Perl 5.10+, 13 + 1 = 14 chars

say!($_&$_/3)

Uses the same method from an old FWP thread as my GolfScript entry. Prints 1 if the input is a power of two, and an empty line otherwise.

Needs to be run with perl -nE; the n costs one extra char, for a total of 14 chars. Alternatively, here's an 18-character version that doesn't need the n:

say!(($_=<>)&$_/3)

python 3, 38

print(1==bin(int(input())).count('1'))

python, 32

However, the code doesn't work in every version.

print 1==bin(input()).count('1')

Notice that the solution works also for 0 (print False).

GTB, 46 bytes

2→P`N[@N=Pg;1P^2→P@P>1E90g;2]l;2~"NO"&l;1"YES"

Python, 35

print bin(input()).strip('0')=='b1'

Doesn't use not only +/- operations, but any math operations aside from converting to binary form.

Other stuff (interesting, but not for competition):

I have also a regexp version (61):

import re;print re.match(r'^0b10+$',bin(input())) is not None

(Love the idea, but import and match function make it too long)

And nice, but boring bitwise operations version (31):

x=input();print x and not x&~-x

(yes, it's shorter, but it uses ~-x for decrement which comtains - operation)

Python 2.7 (30 29 39 37)

EDIT: Updated due to user input requirement;

a=input()
while a>1:a/=2.
print a==1

Brute force, try to divide until =1 (success) or <1 (fail)

JavaScript, 35

Works for bytes.

alert((+prompt()).toString(2)%9==1)

46 character version, Works for 16 bit numbers.

x=(+prompt()).toString(2)%99;alert(x==1|x==10)

The trick works in most dynamic languages.

Explanation: Convert the number to base 2, interpret that string as base 10, do modulo 9 to get the digit sum, which must be 1.

Python (33)

print int(bin(input())[3:]or 0)<1

Ruby, 33, 28, 25

p /.1/!~gets.to_i.to_s(2)

Python, 31

print 3>bin(input()).rfind('1')

APL (12 for 0/1, 27 for yes/no)

≠/A=2*0,ιA←⍞ 

or, if we must output text:

3↑(3x≠/A=2*0,ιA←⍞)↓'YESNO '

Read in A. Form a vector 0..A, then a vector 2⁰..2^A (yes, that's way more than necessary), then a vector comparing A with each of those (resulting in a vector of 0's and at most one 1), then xor that (there's no xor operator in APL, but ≠ applied to booleans will act as one.) We now have 0 or 1.

To get YES or NO: multiply the 0 or 1 by 3, drop this number of characters from 'YESNO ', then take the first 3 characters of this.

C, 65 bytes

main(k){scanf("%i",&k);while(k&&!(k%2))k/=2;puts(k==1?"T":"F");}

C, 48

main(x){scanf("%i",&x);puts(x&~(x*~0)?"F":"T");}

Ruby — 17 characters (fourth try)

p /.1/!~'%b'%gets

My current best is a fusion of @steenslag's answer with my own. Below are my previous attempts.

Ruby — 19 characters (third try)

p /10*1/!~'%b'%gets

Ruby — 22 characters (second try)

p !('%b'%gets)[/10*1/]

Ruby — 24 characters (first try)

p !!('%b'%gets=~/^10*$/)

C# (56 characters)

 Math.Log(Int32.Parse(Console.ReadLine()),2)%1==0?"Y":"N"

PHP, 86 chars

function P($a,$c){if($c==$a)return 1;if($c>$a)return 0;return P($a,$c*2);}echo P(x,2);

Replace x with the number you want to test.

PHP (58)

<?php $a=fgets(STDIN);while($a%2==0)$a/=2;echo $a==1?1:0;

Simple divide-by-2 loop and remainder check.

ActionScript 3 (63 characters*)

var r = Math.log(prompt())/Math.log(2);
return ( (r > 0) && (int(r) == r) );

* I'm going to ignore the fact that AS3 doesn't have a built-in prompt() method, (and hope AS3's lack of implementation doesn't get counted against me).

Extra white-space left in for readability and clarity, but not counted, since they can be removed before execution.

GTB, 17

With 1/0

`Ar?A>1:A/2→A~A=1

With YES/NO (31)

`Ar?A>1:A/2→A@A=1$~"YES"#~"NO"&