If (2^x+2^y)^2=2^m+2^n, where x,y,m and n are all positive integers, then:?

Question

a. x and y are consecutive numbers
b. the different between x and y is 2
c. both x and y are even numbers
d. both x and y are odd numbers
e. both x and y are greater than 4

Answer 1

If x and y are consecutive integers then this identity works:

(2^x + 2^(x + 1))^2 = 2^2x (1 + 2)^2 = 2^2x * 9 = 2^2x + 2^(2x + 3) so m = 2x and n = 2x + 3

However proving the converse is quite difficult.

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Answer 2

If x=y then we have:
2^(2x+2) = 2^m+2^n
Now the only time a sum of (positive) powers of two can be a power of 2 is if both summands are the same power of 2. Thus m=n:
2^(2x+2) = 2^(m+1)
Hence,
2x + 2 = m + 1
or
m = n = 2x+1
Examples:
x=y=1 => m=n=3: (2+2)² = 2³ + 2³
x=y=2 => m=n=5: (2²+2²)² = 2^5 + 2^5


Now assume that y and x are distinct. WLOG (that means without loss of generality), assume WLOG that x<=y and m<=n.

Then expand the original equation and factor out a 2^(2x) on the left and a 2^m on the right (assume that m<=n):
(2^(2x)) (1 + 2^(2y-2x) + 2^(y-x+1)) = (2^m) (1+2^(n-m))

Now the right part of the left side is odd and greater than one, whereas the left part of the left side is a(n even) power of 2.

If m were to equal n, then the right side would be a power of two, without any odd factor greater than 1, a contradiction. Thus m is not equal to n.

So assume (WLOG) that m 2x = m

Similarly, the right part of both sides must be equal:
1 + 2^(2y-2x) + 2^(y-x+1) = 1+2^(n-m) or
2^(2y-2x) + 2^(y-x+1) = 2^(n-m)
Once again, the only way that the left side (summing two powers of 2) can be a power of two is for both of the summands to be equal:
2y-2x = y-x+1 or
y = x + 1
Thus the left side is 8 so that n-m = 3

Therefore,
y = x+1
m = 2x
n = 2x + 3

Examples:
x=1, y=2 => m=2, n=5: (2+2²)² = 2² + 2^5 = 36
x=2, y=3 => m=4, n=7: (2²+2³)² = 2^4 + 2^7 = 144


In conclusion (and combining the two cases),
y is either x or x+1
m = 3x - y +1 and
n = 2y + 1

Notice that this means that none of a through e are necessarily true. And b is definitely false.

Answer 3

It is easy to show that (a) works, i.e. x and y could be consecutive numbers. Whether they have to be I'm not sure. It is also easy to show that (e) is false as x = 1 and y = 2 works all right giving n = 5 and m = 2 (or vice versa) and this of course proves that (c) and (d) are false.
I haven't worked on it long enough to be sure of (b) but I think that this will turn out to be false. Were you given that only one answer was correct?