How to prove that either $2^{500} + 15$ or $2^{500} + 16$ isn't a perfect square? How would I prove that either $2^{500} + 15$ or $2^{500} + 16$ isn't a perfect square?
 A: First write down exactly what it is that you're trying to prove is impossible.  In this case, that means writing down these two equations:
$$2^{500}+15=n^2$$
$$2^{500}+16=m^2$$
Next, play with the equations and see where they lead you.  If you subtract the first from the second, you get
$$1=m^2-n^2$$
or 
$$1=(m+n)(m-n)$$
Now think about what this tells you about $m$ and $n$.  Hint:  What are all the factors of $1$?  
A: The question is not about that big numbers...
Question is about a result that two consecutive numbers can not be squares simultaneously..
Suppose  $a=b^2\text { and }a+1=c^2\Rightarrow b^2+1=c^2\Rightarrow c^2-b^2=1\Rightarrow (c+b)(c-b)=1$
w.l.o.g. assume $c+b=1$ which implies 
$a+1=c^2=(1-b)^2=1+b^2-2b\Rightarrow a=b^2-2b$ but then we have $a=b^2$
you should now be able to see some contradiction...
So...
A: Well, it's obvious isn't it?
How could two consecutive numbers of that size be squares? Write $a = 2^{500}+15$  and $2^{500}+16=a+1$. Imagine $a$ were a square. Then its square root would be a pretty huge number. If you so much as add $1$ to that number, the square of the result will be massively larger than $a$, there'll be no way to get something that squares to $a+1$.
That's the gist of the other answers. The first step in many proofs is to think about why the answer is obvious.
A: Alternative method (for fun):
Observe $2^4 = 16 \equiv 1$ mod $5$.
Then $2^{500} = (2^4)^{125} \equiv 1^{125} = 1$ mod $5$.
Thus, $2^{500} + 16 \equiv 1 + 1 = 2$ mod $5$.
But every square is either $0, 1$ or $4$ mod $5$.
Therefore, $2^{500} + 16$ cannot be a square. QED
A: Hint: think about the distance between consecutive squares. 
A: Short solutions: $2^{1000} + 15 \equiv 3 \pmod{4}$, $2^{1000} + 16 \equiv 2 \pmod{3}$, neither of which are quadratic residues.
More elementary solution: $2^{1000} = (2^{500})^2$. The next perfect square is $$(2^{500} + 1)^2 = 2^{1000} + 2(2^{500}) + 1$$ which is clearly more than both $2^{1000} + 15$ and $2^{1000} + 16$. Therefore, these two can't possibly be perfect squares.
A: To expand on Martin's hint a bit:
The difference between two perfect squares $n^2$ and $(n+1)^2$ is
$$(n+1)^2 - n^2 = 2n + 1.$$
You have two numbers that are expressed as the sum of a perfect square and another number.
What would be the smallest value that the other number would have to be in order for the sum to be a perfect square also?
A: Oh well, $2^{500}$ is a perfect square, namely it is $(2^{250})^2$. Any perfect square bigger than $2^{500}$ is at least $(2^{250}+1)^2$, that is $2^{500} + 2\cdot2^{250}+1$, which is of course much bigger than $2^{500}+16$.
A: All square numbers, modulo 16, are in $\{0, 1, 4, 9\}$.  But $2^{500} + 15 = 16^{125} + 15 \equiv 15$ (mod 16), so it can't be a square number.
A: $2^{500} + 15$ is not a perfect square:


*

*$2^{500} \equiv 0 \pmod 4$

*$15    \equiv 3 \pmod 4$

*$2^{500} + 15 \equiv 3 \pmod 4$


A perfect square is congruent to 0 or 1 modulo 4,
so $2^{500} + 15$ is not a perfect square.
