Method 1:lime777 wrote:Find, with justification, all positive integers n such that 2^2 + 2^5 + 2^n is a perfect square.
(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
