When is $32x+32$ a square number? I am trying to find out for what values for $x$ does the function $f(x)=32x+32$ return a square number? I found that this is the case for at least: $x \in \{64,1600,53824,1827904,62094400,2109381184\}$ but I can't see a pattern in those numbers.
Is there a formula which generates these values for $x$ instead of simply trying a lot of values?
 A: The answer by user3491648 gives a complete characterization of the full set of values $x$ for which $32x+32$ is a square numbers.  This is really just an observation on the set of values 
$$\{64,1600,53824,1827904,62094400,2109381184\}$$
reported by the OP.  
These are not actually values that $x$ takes on.  (As user3491648 finds, it requires on odd value of $x$ to make $32x+32$ a square.)  Rather, these seem to be the square values that correspond to values of $x$ that are themselves squares, namely of the following values:
$$\{1,7,41,239,1393,8119\}$$
For example, $32\cdot239^2+32=1827904$.  It's unclear why these values are being singled out, but they do seem to obey a two-term recursion, namely
$$a_{n+1}=6a_n-a_{n-1}$$
For example, $1393=6\cdot239-41$.
A: $$32(x + 1) = y^2 \implies 32 \mid y^2 \implies 4 \mid y$$ so let $y = 4z$. $$2(x+1) = z^2 \implies 2 \mid z$$ so let $z = 2w$ $$x + 1 = 2w^2 $$ so given any $w$ we can solve for $x$.
So the set of solutions is
$$\{(2w^2 - 1, 8w) \mid w \in \Bbb Z\}$$
A: Solution for x such that 32x+32 is a square as follows:
 For integral solution,  x= (n2 -2)/2 such that x is in Natural number.(n is in natural number)
