Tough problem

lime777 wrote:Find, with justification, all positive integers n such that 2^2 + 2^5 + 2^n is a perfect square.

Method 1:
(2^2 + 2^5 + 2^n) = (4 + 32 + 2^n) = (36 + 2^n)
Note that (36 + 2^n) being a multiple of 2, cannot be a perfect square of an odd integer.
Hence, if it is a perfect square, it must be a square of an even integer.

Say, the even integer is 2m.
So, (36 + 2^n) = (2m)^2
--> 2^n = (2m)^2 - 36 = (2m)^2 - 6^2 = (2m - 6)(2m + 6)

Now, (2m - 6) and (2m + 6) both must be some power of 2 only as their product is 2^n.

Say, (2m - 6) = 2^x and (2m + 6) = 2^y
Hence, (2^y - 2^x) = 12
Now, 12 can be expressed as the difference of two powers of 2 in only one way, (2^4 - 2^2) = (16 - 4) = 12

Hence, there is only one possible set of values for x and y, x = 2 and y = 4 and thus only one possible value of n.

Hence, 2^n = (2^x)*(2^y) = (2^2)*(2^4) = 2^6
Hence, n = 6

Method 2:
Any perfect square can be written in the form (a + b)^2 = (a^2 + 2ab + b^2)
Hence, we'll try to write (2^2 + 2^5 + 2^n) in such manner.

Now, if we take a^2 = 2^2 and 2ab = 2^5, then b = 2^3 = 8

Hence, (2^2 + 2*2*8 + 8^2) = (2 + 8)^2 = 10^2

Hence, 2^n = 8^2 = 2^6
Hence, n = 6

Now this is the only possible way to express (2^2 + 2^5 + 2^n) as (a + b)^2 such (a + b) is an integer. Hence, only possible positive integral value of n is 6.