More fun in 2016, Part 13

$\dfrac{n^2+32^2}{n+32}=\dfrac{n^2-32^2}{n+32}+\dfrac{2048}{n+32}=n-32+\dfrac{2048}{n+32}$ $n+32|2048$ $(n+32)_{\text{max}}=2048$ $n_{\text{max}}=\boxed{2016}$

$n^2+32^2=(n+32)^2-64n$

thus, $64n$ should be divisible by $n+32$

$64n=(n+32)k$ where k is a positive integer.

$\therefore n= \frac{32k}{64-k}$

Now n is max. when $64-k$ is min. i.e. 1

$\therefore k=63, n=32*63=2016$ .

Divide n^2+32 by n+32, we'll get a remainder 2(32)^2, then we'll have the equation 2(32^2)/(n+32)=1 [we let the expression be equal to 1 so that we can get the maximum integer value of n].. We have then n=2016..

Hint: Use algebraic substitution and let u=n+32